3 The distribution of the weights of cereal in boxes of a specific brand of cereal...

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3 The distribution of the weights of cereal in boxes of a specific brand of cereal is approximately normally distributed with mean 13 ounces and standard deviation 0.12 ounce. Complete parts (a) and (b) below. CHILE a. Explain why it would be unwise for the cereal company to advertise that each box contains 13 ounces of cereal. OA. Only half of the boxes of cereal would have exactly 13 ounces of cereal. OB. Less than half of the boxes of cereal would have at least 13 ounces of cereal. OC. More than half of the boxes of cereal would have at least 13 ounces of cereal. O D. Only half of the boxes of cereal would have at least 13 ounces of cereal. b. What should the company advertise the weight of the cereal in each box to be so that 99% of the boxes contain at least that much cereal? The company should advertise the weight of the cereal in each box to be 12.6 ounces. (Round to one decimal place as needed.) Time Remaining: 00:05:44 Next For students who graduated from high school in 2013, their scores on a standardized math test are approximately normally distributed with a mean of 505 points and a standard = deviation of 121 points. Complete parts (a) through (c) below. OA. Yes. The student's score is unusual. The associated z-score is OB. Yes. The student's score is unusual. The associated z-score is OC. No. The student's score is not unusual. The associated z-score is OD. No. The student's score is not unusual. The associated z-score is THR which is greater than the positive z-score required to determine whether an observation is unusual. which is less than the positive z-score required to determine whether an observation is unusual. .96 which is less than the positive z-score required to determine whether an observation is unusual. which is greater than the positive z-score required to determine whether an observation is unusual. b. A student who graduated from high school in 2013 scored 787 points on a standardized math test. Is that an unusual score? Explain. Select the correct choice below and fill in the answer box to complete your choice. (Round to two decimal places as needed.) OA. Yes. The student's score is unusual. The associated z-score is OB. No. The student's score is not unusual. The associated z-score is OC. No. The student's score is not unusual. The associated z-score is OD. Yes. The student's score is unusual. The associated z-score is 99 which is greater than the positive z-score required to determine whether an observation is unusual. which is less than the positive z-score required to determine whether an observation is unusual. which is greater than the positive z-score required to determine whether an observation is unusual. 99 which is less than the positive z-score required to determine whether an observation is unusual. c. For students who graduated from high school in 2013, find the cutoff for unusually high scores on a standardized math test. cording 21.04 PM Condo ead.xlsx cording 23.58 PM (a) throu (c) below. UB. Yes. The student's score is unusual. The associated z-score is OC. No. The student's score is not unusual. The associated z-score is OD. No. The student's score is not unusual. The associated z-score is TE which is less than the positive z-score required to determine whether an observation is unusual. 96 which is less than the positive z-score required to determine whether an observation is unusual. which is greater than the positive z-score required to determine whether an observation is unusual. b. A student who graduated from high school in 2013 scored 787 points on a standardized math test. Is that an unusual score? Explain. Select the correct choice below and fill in the answer box to complete your choice. (Round to two decimal places as needed.) OA. Yes. The student's score is unusual. The associated z-score is OB. No. The student's score is not unusual. The associated z-score is OC. No. The student's score is not unusual. The associated z-score is which is less than the positive z-score required to determine whether an observation is unusual. which is greater than the positive z-score required to determine whether an observation is unusual. 99 which is less than the positive z-score required to determine whether an observation is unusual. D. Yes. The student's score is unusual. The associated z-score is .99 which is greater than the positive z-score required to determine whether an observation is unusual. c. For students who graduated from high school in 2013, find the cutoff for unusually high scores on a standardized math test. 747 points (Round up to the nearest whole number.) corc Condo ead.xls cording 23.58 PM N K Without calculating P(Z < 1.3), sketch the standard normal curve to help you explain why P(Z < 1.3) cannot be equal to 0.3. Sketch the standard normal curve with the desired area shaded. Choose the correct graph below. O A. in 3-2-1 0 1 2 3 Density O Q OB. Density 0- -3-2-1 0 1 2 3 Q Q G Why can P(Z <1.3) not be equal to 0.3? OA. The area of the shaded region is greater than 0.5, so it cannot be 0.3. OB. The area of the shaded region is greater than 1, so it cannot be 0.3. OC. The area of the shaded region is less than 0.2, so it cannot be 0.3. OD. The area of the shaded region is less than 0.3, so it cannot be 0.3. Density C. 0- -3-2-1 0 1 Q M Density D. -3-2-1 0 1 2 3 Skry 6 77 K Find the probability that a z-score randomly selected from the standard normal distribution meets the given condition. Use probability notation to describe your result. The Z-score is at least 1.12. Click here to view page 1 of the standard normal distribution table. Click here to view page 2 of the standard normal distribution table. The probability is P(Z≥1.12) = 8686 (Round to four decimal places as needed.) 1 K In a certain region, women's heights are approximately normally distributed with mean 63.8 inches and standard deviation 4.1 inches. Men's heights are approximately normally distributed with mean 69.5 inches and standard deviation 4.6 inches. Complete parts (a) through (d) below. Click here to view page 1 of the standard normal distribution table. Click here to view page 2 of the standard normal distribution table. a. At what doorway height would 95% of men be able to pass through without having to duck? www 78.7 inches (Round to the nearest whole number as needed.) b. What percentage of women would have to duck to pass through a doorway at the height found in part (a)? .03% (Round to two decimal places as needed.) c. The height of a doorway is typically 80 inches. What percentage of men can pass through such a doorway without ducking their head? 28% (Round to two decimal places as needed.) d. The mean height of men and the mean height of women have both increased over time. For each of parts (a), (b), and (c), determine whether the correct answer to the question in the future will be larger, smaller, or impossible to know from the given information. The answer to part (a) would the answer to part (b) would increase, decrease, and the answer to part (c) would be impossible to determine. K Find the probability that a z-score randomly selected from the standard normal distribution meets the given condition. The Z-score is less than 0.85. Click to view page 1 of the table. Click to view page 2 of the table. The probability is .8023 (Round to four decimal places as needed.) K The lengths of songs played on one day on a radio station were analyzed. The lengths of the songs are approximately normally distributed with mean 249.21 seconds and standar deviation 44.54 seconds. Find the z-score of a song that is 280 seconds long. What does it mean in this situation? CI The Z-score is 69. This means that the song is 31 standard deviations longer than the mean. (Round to two decimal places as needed.) 3 The distribution of the weights of cereal in boxes of a specific brand of cereal is approximately normally distributed with mean 13 ounces and standard deviation 0.12 ounce. Complete parts (a) and (b) below. CHILE a. Explain why it would be unwise for the cereal company to advertise that each box contains 13 ounces of cereal. OA. Only half of the boxes of cereal would have exactly 13 ounces of cereal. OB. Less than half of the boxes of cereal would have at least 13 ounces of cereal. OC. More than half of the boxes of cereal would have at least 13 ounces of cereal. O D. Only half of the boxes of cereal would have at least 13 ounces of cereal. b. What should the company advertise the weight of the cereal in each box to be so that 99% of the boxes contain at least that much cereal? The company should advertise the weight of the cereal in each box to be 12.6 ounces. (Round to one decimal place as needed.) Time Remaining: 00:05:44 Next For students who graduated from high school in 2013, their scores on a standardized math test are approximately normally distributed with a mean of 505 points and a standard = deviation of 121 points. Complete parts (a) through (c) below. OA. Yes. The student's score is unusual. The associated z-score is OB. Yes. The student's score is unusual. The associated z-score is OC. No. The student's score is not unusual. The associated z-score is OD. No. The student's score is not unusual. The associated z-score is THR which is greater than the positive z-score required to determine whether an observation is unusual. which is less than the positive z-score required to determine whether an observation is unusual. .96 which is less than the positive z-score required to determine whether an observation is unusual. which is greater than the positive z-score required to determine whether an observation is unusual. b. A student who graduated from high school in 2013 scored 787 points on a standardized math test. Is that an unusual score? Explain. Select the correct choice below and fill in the answer box to complete your choice. (Round to two decimal places as needed.) OA. Yes. The student's score is unusual. The associated z-score is OB. No. The student's score is not unusual. The associated z-score is OC. No. The student's score is not unusual. The associated z-score is OD. Yes. The student's score is unusual. The associated z-score is 99 which is greater than the positive z-score required to determine whether an observation is unusual. which is less than the positive z-score required to determine whether an observation is unusual. which is greater than the positive z-score required to determine whether an observation is unusual. 99 which is less than the positive z-score required to determine whether an observation is unusual. c. For students who graduated from high school in 2013, find the cutoff for unusually high scores on a standardized math test. cording 21.04 PM Condo ead.xlsx cording 23.58 PM (a) throu (c) below. UB. Yes. The student's score is unusual. The associated z-score is OC. No. The student's score is not unusual. The associated z-score is OD. No. The student's score is not unusual. The associated z-score is TE which is less than the positive z-score required to determine whether an observation is unusual. 96 which is less than the positive z-score required to determine whether an observation is unusual. which is greater than the positive z-score required to determine whether an observation is unusual. b. A student who graduated from high school in 2013 scored 787 points on a standardized math test. Is that an unusual score? Explain. Select the correct choice below and fill in the answer box to complete your choice. (Round to two decimal places as needed.) OA. Yes. The student's score is unusual. The associated z-score is OB. No. The student's score is not unusual. The associated z-score is OC. No. The student's score is not unusual. The associated z-score is which is less than the positive z-score required to determine whether an observation is unusual. which is greater than the positive z-score required to determine whether an observation is unusual. 99 which is less than the positive z-score required to determine whether an observation is unusual. D. Yes. The student's score is unusual. The associated z-score is .99 which is greater than the positive z-score required to determine whether an observation is unusual. c. For students who graduated from high school in 2013, find the cutoff for unusually high scores on a standardized math test. 747 points (Round up to the nearest whole number.) corc Condo ead.xls cording 23.58 PM N K Without calculating P(Z < 1.3), sketch the standard normal curve to help you explain why P(Z < 1.3) cannot be equal to 0.3. Sketch the standard normal curve with the desired area shaded. Choose the correct graph below. O A. in 3-2-1 0 1 2 3 Density O Q OB. Density 0- -3-2-1 0 1 2 3 Q Q G Why can P(Z <1.3) not be equal to 0.3? OA. The area of the shaded region is greater than 0.5, so it cannot be 0.3. OB. The area of the shaded region is greater than 1, so it cannot be 0.3. OC. The area of the shaded region is less than 0.2, so it cannot be 0.3. OD. The area of the shaded region is less than 0.3, so it cannot be 0.3. Density C. 0- -3-2-1 0 1 Q M Density D. -3-2-1 0 1 2 3 Skry 6 77 K Find the probability that a z-score randomly selected from the standard normal distribution meets the given condition. Use probability notation to describe your result. The Z-score is at least 1.12. Click here to view page 1 of the standard normal distribution table. Click here to view page 2 of the standard normal distribution table. The probability is P(Z≥1.12) = 8686 (Round to four decimal places as needed.) 1 K In a certain region, women's heights are approximately normally distributed with mean 63.8 inches and standard deviation 4.1 inches. Men's heights are approximately normally distributed with mean 69.5 inches and standard deviation 4.6 inches. Complete parts (a) through (d) below. Click here to view page 1 of the standard normal distribution table. Click here to view page 2 of the standard normal distribution table. a. At what doorway height would 95% of men be able to pass through without having to duck? www 78.7 inches (Round to the nearest whole number as needed.) b. What percentage of women would have to duck to pass through a doorway at the height found in part (a)? .03% (Round to two decimal places as needed.) c. The height of a doorway is typically 80 inches. What percentage of men can pass through such a doorway without ducking their head? 28% (Round to two decimal places as needed.) d. The mean height of men and the mean height of women have both increased over time. For each of parts (a), (b), and (c), determine whether the correct answer to the question in the future will be larger, smaller, or impossible to know from the given information. The answer to part (a) would the answer to part (b) would increase, decrease, and the answer to part (c) would be impossible to determine. K Find the probability that a z-score randomly selected from the standard normal distribution meets the given condition. The Z-score is less than 0.85. Click to view page 1 of the table. Click to view page 2 of the table. The probability is .8023 (Round to four decimal places as needed.) K The lengths of songs played on one day on a radio station were analyzed. The lengths of the songs are approximately normally distributed with mean 249.21 seconds and standar deviation 44.54 seconds. Find the z-score of a song that is 280 seconds long. What does it mean in this situation? CI The Z-score is 69. This means that the song is 31 standard deviations longer than the mean. (Round to two decimal places as needed.) 3 The distribution of the weights of cereal in boxes of a specific brand of cereal is approximately normally distributed with mean 13 ounces and standard deviation 0.12 ounce. Complete parts (a) and (b) below. CHILE a. Explain why it would be unwise for the cereal company to advertise that each box contains 13 ounces of cereal. OA. Only half of the boxes of cereal would have exactly 13 ounces of cereal. OB. Less than half of the boxes of cereal would have at least 13 ounces of cereal. OC. More than half of the boxes of cereal would have at least 13 ounces of cereal. O D. Only half of the boxes of cereal would have at least 13 ounces of cereal. b. What should the company advertise the weight of the cereal in each box to be so that 99% of the boxes contain at least that much cereal? The company should advertise the weight of the cereal in each box to be 12.6 ounces. (Round to one decimal place as needed.) Time Remaining: 00:05:44 Next For students who graduated from high school in 2013, their scores on a standardized math test are approximately normally distributed with a mean of 505 points and a standard = deviation of 121 points. Complete parts (a) through (c) below. OA. Yes. The student's score is unusual. The associated z-score is OB. Yes. The student's score is unusual. The associated z-score is OC. No. The student's score is not unusual. The associated z-score is OD. No. The student's score is not unusual. The associated z-score is THR which is greater than the positive z-score required to determine whether an observation is unusual. which is less than the positive z-score required to determine whether an observation is unusual. .96 which is less than the positive z-score required to determine whether an observation is unusual. which is greater than the positive z-score required to determine whether an observation is unusual. b. A student who graduated from high school in 2013 scored 787 points on a standardized math test. Is that an unusual score? Explain. Select the correct choice below and fill in the answer box to complete your choice. (Round to two decimal places as needed.) OA. Yes. The student's score is unusual. The associated z-score is OB. No. The student's score is not unusual. The associated z-score is OC. No. The student's score is not unusual. The associated z-score is OD. Yes. The student's score is unusual. The associated z-score is 99 which is greater than the positive z-score required to determine whether an observation is unusual. which is less than the positive z-score required to determine whether an observation is unusual. which is greater than the positive z-score required to determine whether an observation is unusual. 99 which is less than the positive z-score required to determine whether an observation is unusual. c. For students who graduated from high school in 2013, find the cutoff for unusually high scores on a standardized math test. cording 21.04 PM Condo ead.xlsx cording 23.58 PM (a) throu (c) below. UB. Yes. The student's score is unusual. The associated z-score is OC. No. The student's score is not unusual. The associated z-score is OD. No. The student's score is not unusual. The associated z-score is TE which is less than the positive z-score required to determine whether an observation is unusual. 96 which is less than the positive z-score required to determine whether an observation is unusual. which is greater than the positive z-score required to determine whether an observation is unusual. b. A student who graduated from high school in 2013 scored 787 points on a standardized math test. Is that an unusual score? Explain. Select the correct choice below and fill in the answer box to complete your choice. (Round to two decimal places as needed.) OA. Yes. The student's score is unusual. The associated z-score is OB. No. The student's score is not unusual. The associated z-score is OC. No. The student's score is not unusual. The associated z-score is which is less than the positive z-score required to determine whether an observation is unusual. which is greater than the positive z-score required to determine whether an observation is unusual. 99 which is less than the positive z-score required to determine whether an observation is unusual. D. Yes. The student's score is unusual. The associated z-score is .99 which is greater than the positive z-score required to determine whether an observation is unusual. c. For students who graduated from high school in 2013, find the cutoff for unusually high scores on a standardized math test. 747 points (Round up to the nearest whole number.) corc Condo ead.xls cording 23.58 PM N K Without calculating P(Z < 1.3), sketch the standard normal curve to help you explain why P(Z < 1.3) cannot be equal to 0.3. Sketch the standard normal curve with the desired area shaded. Choose the correct graph below. O A. in 3-2-1 0 1 2 3 Density O Q OB. Density 0- -3-2-1 0 1 2 3 Q Q G Why can P(Z <1.3) not be equal to 0.3? OA. The area of the shaded region is greater than 0.5, so it cannot be 0.3. OB. The area of the shaded region is greater than 1, so it cannot be 0.3. OC. The area of the shaded region is less than 0.2, so it cannot be 0.3. OD. The area of the shaded region is less than 0.3, so it cannot be 0.3. Density C. 0- -3-2-1 0 1 Q M Density D. -3-2-1 0 1 2 3 Skry 6 77 K Find the probability that a z-score randomly selected from the standard normal distribution meets the given condition. Use probability notation to describe your result. The Z-score is at least 1.12. Click here to view page 1 of the standard normal distribution table. Click here to view page 2 of the standard normal distribution table. The probability is P(Z≥1.12) = 8686 (Round to four decimal places as needed.) 1 K In a certain region, women's heights are approximately normally distributed with mean 63.8 inches and standard deviation 4.1 inches. Men's heights are approximately normally distributed with mean 69.5 inches and standard deviation 4.6 inches. Complete parts (a) through (d) below. Click here to view page 1 of the standard normal distribution table. Click here to view page 2 of the standard normal distribution table. a. At what doorway height would 95% of men be able to pass through without having to duck? www 78.7 inches (Round to the nearest whole number as needed.) b. What percentage of women would have to duck to pass through a doorway at the height found in part (a)? .03% (Round to two decimal places as needed.) c. The height of a doorway is typically 80 inches. What percentage of men can pass through such a doorway without ducking their head? 28% (Round to two decimal places as needed.) d. The mean height of men and the mean height of women have both increased over time. For each of parts (a), (b), and (c), determine whether the correct answer to the question in the future will be larger, smaller, or impossible to know from the given information. The answer to part (a) would the answer to part (b) would increase, decrease, and the answer to part (c) would be impossible to determine. K Find the probability that a z-score randomly selected from the standard normal distribution meets the given condition. The Z-score is less than 0.85. Click to view page 1 of the table. Click to view page 2 of the table. The probability is .8023 (Round to four decimal places as needed.) K The lengths of songs played on one day on a radio station were analyzed. The lengths of the songs are approximately normally distributed with mean 249.21 seconds and standar deviation 44.54 seconds. Find the z-score of a song that is 280 seconds long. What does it mean in this situation? CI The Z-score is 69. This means that the song is 31 standard deviations longer than the mean. (Round to two decimal places as needed.) 3 The distribution of the weights of cereal in boxes of a specific brand of cereal is approximately normally distributed with mean 13 ounces and standard deviation 0.12 ounce. Complete parts (a) and (b) below. CHILE a. Explain why it would be unwise for the cereal company to advertise that each box contains 13 ounces of cereal. OA. Only half of the boxes of cereal would have exactly 13 ounces of cereal. OB. Less than half of the boxes of cereal would have at least 13 ounces of cereal. OC. More than half of the boxes of cereal would have at least 13 ounces of cereal. O D. Only half of the boxes of cereal would have at least 13 ounces of cereal. b. What should the company advertise the weight of the cereal in each box to be so that 99% of the boxes contain at least that much cereal? The company should advertise the weight of the cereal in each box to be 12.6 ounces. (Round to one decimal place as needed.) Time Remaining: 00:05:44 Next For students who graduated from high school in 2013, their scores on a standardized math test are approximately normally distributed with a mean of 505 points and a standard = deviation of 121 points. Complete parts (a) through (c) below. OA. Yes. The student's score is unusual. The associated z-score is OB. Yes. The student's score is unusual. The associated z-score is OC. No. The student's score is not unusual. The associated z-score is OD. No. The student's score is not unusual. The associated z-score is THR which is greater than the positive z-score required to determine whether an observation is unusual. which is less than the positive z-score required to determine whether an observation is unusual. .96 which is less than the positive z-score required to determine whether an observation is unusual. which is greater than the positive z-score required to determine whether an observation is unusual. b. A student who graduated from high school in 2013 scored 787 points on a standardized math test. Is that an unusual score? Explain. Select the correct choice below and fill in the answer box to complete your choice. (Round to two decimal places as needed.) OA. Yes. The student's score is unusual. The associated z-score is OB. No. The student's score is not unusual. The associated z-score is OC. No. The student's score is not unusual. The associated z-score is OD. Yes. The student's score is unusual. The associated z-score is 99 which is greater than the positive z-score required to determine whether an observation is unusual. which is less than the positive z-score required to determine whether an observation is unusual. which is greater than the positive z-score required to determine whether an observation is unusual. 99 which is less than the positive z-score required to determine whether an observation is unusual. c. For students who graduated from high school in 2013, find the cutoff for unusually high scores on a standardized math test. cording 21.04 PM Condo ead.xlsx cording 23.58 PM (a) throu (c) below. UB. Yes. The student's score is unusual. The associated z-score is OC. No. The student's score is not unusual. The associated z-score is OD. No. The student's score is not unusual. The associated z-score is TE which is less than the positive z-score required to determine whether an observation is unusual. 96 which is less than the positive z-score required to determine whether an observation is unusual. which is greater than the positive z-score required to determine whether an observation is unusual. b. A student who graduated from high school in 2013 scored 787 points on a standardized math test. Is that an unusual score? Explain. Select the correct choice below and fill in the answer box to complete your choice. (Round to two decimal places as needed.) OA. Yes. The student's score is unusual. The associated z-score is OB. No. The student's score is not unusual. The associated z-score is OC. No. The student's score is not unusual. The associated z-score is which is less than the positive z-score required to determine whether an observation is unusual. which is greater than the positive z-score required to determine whether an observation is unusual. 99 which is less than the positive z-score required to determine whether an observation is unusual. D. Yes. The student's score is unusual. The associated z-score is .99 which is greater than the positive z-score required to determine whether an observation is unusual. c. For students who graduated from high school in 2013, find the cutoff for unusually high scores on a standardized math test. 747 points (Round up to the nearest whole number.) corc Condo ead.xls cording 23.58 PM N K Without calculating P(Z < 1.3), sketch the standard normal curve to help you explain why P(Z < 1.3) cannot be equal to 0.3. Sketch the standard normal curve with the desired area shaded. Choose the correct graph below. O A. in 3-2-1 0 1 2 3 Density O Q OB. Density 0- -3-2-1 0 1 2 3 Q Q G Why can P(Z <1.3) not be equal to 0.3? OA. The area of the shaded region is greater than 0.5, so it cannot be 0.3. OB. The area of the shaded region is greater than 1, so it cannot be 0.3. OC. The area of the shaded region is less than 0.2, so it cannot be 0.3. OD. The area of the shaded region is less than 0.3, so it cannot be 0.3. Density C. 0- -3-2-1 0 1 Q M Density D. -3-2-1 0 1 2 3 Skry 6 77 K Find the probability that a z-score randomly selected from the standard normal distribution meets the given condition. Use probability notation to describe your result. The Z-score is at least 1.12. Click here to view page 1 of the standard normal distribution table. Click here to view page 2 of the standard normal distribution table. The probability is P(Z≥1.12) = 8686 (Round to four decimal places as needed.) 1 K In a certain region, women's heights are approximately normally distributed with mean 63.8 inches and standard deviation 4.1 inches. Men's heights are approximately normally distributed with mean 69.5 inches and standard deviation 4.6 inches. Complete parts (a) through (d) below. Click here to view page 1 of the standard normal distribution table. Click here to view page 2 of the standard normal distribution table. a. At what doorway height would 95% of men be able to pass through without having to duck? www 78.7 inches (Round to the nearest whole number as needed.) b. What percentage of women would have to duck to pass through a doorway at the height found in part (a)? .03% (Round to two decimal places as needed.) c. The height of a doorway is typically 80 inches. What percentage of men can pass through such a doorway without ducking their head? 28% (Round to two decimal places as needed.) d. The mean height of men and the mean height of women have both increased over time. For each of parts (a), (b), and (c), determine whether the correct answer to the question in the future will be larger, smaller, or impossible to know from the given information. The answer to part (a) would the answer to part (b) would increase, decrease, and the answer to part (c) would be impossible to determine. K Find the probability that a z-score randomly selected from the standard normal distribution meets the given condition. The Z-score is less than 0.85. Click to view page 1 of the table. Click to view page 2 of the table. The probability is .8023 (Round to four decimal places as needed.) K The lengths of songs played on one day on a radio station were analyzed. The lengths of the songs are approximately normally distributed with mean 249.21 seconds and standar deviation 44.54 seconds. Find the z-score of a song that is 280 seconds long. What does it mean in this situation? CI The Z-score is 69. This means that the song is 31 standard deviations longer than the mean. (Round to two decimal places as needed.) 3 The distribution of the weights of cereal in boxes of a specific brand of cereal is approximately normally distributed with mean 13 ounces and standard deviation 0.12 ounce. Complete parts (a) and (b) below. CHILE a. Explain why it would be unwise for the cereal company to advertise that each box contains 13 ounces of cereal. OA. Only half of the boxes of cereal would have exactly 13 ounces of cereal. OB. Less than half of the boxes of cereal would have at least 13 ounces of cereal. OC. More than half of the boxes of cereal would have at least 13 ounces of cereal. O D. Only half of the boxes of cereal would have at least 13 ounces of cereal. b. What should the company advertise the weight of the cereal in each box to be so that 99% of the boxes contain at least that much cereal? The company should advertise the weight of the cereal in each box to be 12.6 ounces. (Round to one decimal place as needed.) Time Remaining: 00:05:44 Next For students who graduated from high school in 2013, their scores on a standardized math test are approximately normally distributed with a mean of 505 points and a standard = deviation of 121 points. Complete parts (a) through (c) below. OA. Yes. The student's score is unusual. The associated z-score is OB. Yes. The student's score is unusual. The associated z-score is OC. No. The student's score is not unusual. The associated z-score is OD. No. The student's score is not unusual. The associated z-score is THR which is greater than the positive z-score required to determine whether an observation is unusual. which is less than the positive z-score required to determine whether an observation is unusual. .96 which is less than the positive z-score required to determine whether an observation is unusual. which is greater than the positive z-score required to determine whether an observation is unusual. b. A student who graduated from high school in 2013 scored 787 points on a standardized math test. Is that an unusual score? Explain. Select the correct choice below and fill in the answer box to complete your choice. (Round to two decimal places as needed.) OA. Yes. The student's score is unusual. The associated z-score is OB. No. The student's score is not unusual. The associated z-score is OC. No. The student's score is not unusual. The associated z-score is OD. Yes. The student's score is unusual. The associated z-score is 99 which is greater than the positive z-score required to determine whether an observation is unusual. which is less than the positive z-score required to determine whether an observation is unusual. which is greater than the positive z-score required to determine whether an observation is unusual. 99 which is less than the positive z-score required to determine whether an observation is unusual. c. For students who graduated from high school in 2013, find the cutoff for unusually high scores on a standardized math test. cording 21.04 PM Condo ead.xlsx cording 23.58 PM (a) throu (c) below. UB. Yes. The student's score is unusual. The associated z-score is OC. No. The student's score is not unusual. The associated z-score is OD. No. The student's score is not unusual. The associated z-score is TE which is less than the positive z-score required to determine whether an observation is unusual. 96 which is less than the positive z-score required to determine whether an observation is unusual. which is greater than the positive z-score required to determine whether an observation is unusual. b. A student who graduated from high school in 2013 scored 787 points on a standardized math test. Is that an unusual score? Explain. Select the correct choice below and fill in the answer box to complete your choice. (Round to two decimal places as needed.) OA. Yes. The student's score is unusual. The associated z-score is OB. No. The student's score is not unusual. The associated z-score is OC. No. The student's score is not unusual. The associated z-score is which is less than the positive z-score required to determine whether an observation is unusual. which is greater than the positive z-score required to determine whether an observation is unusual. 99 which is less than the positive z-score required to determine whether an observation is unusual. D. Yes. The student's score is unusual. The associated z-score is .99 which is greater than the positive z-score required to determine whether an observation is unusual. c. For students who graduated from high school in 2013, find the cutoff for unusually high scores on a standardized math test. 747 points (Round up to the nearest whole number.) corc Condo ead.xls cording 23.58 PM N K Without calculating P(Z < 1.3), sketch the standard normal curve to help you explain why P(Z < 1.3) cannot be equal to 0.3. Sketch the standard normal curve with the desired area shaded. Choose the correct graph below. O A. in 3-2-1 0 1 2 3 Density O Q OB. Density 0- -3-2-1 0 1 2 3 Q Q G Why can P(Z <1.3) not be equal to 0.3? OA. The area of the shaded region is greater than 0.5, so it cannot be 0.3. OB. The area of the shaded region is greater than 1, so it cannot be 0.3. OC. The area of the shaded region is less than 0.2, so it cannot be 0.3. OD. The area of the shaded region is less than 0.3, so it cannot be 0.3. Density C. 0- -3-2-1 0 1 Q M Density D. -3-2-1 0 1 2 3 Skry 6 77 K Find the probability that a z-score randomly selected from the standard normal distribution meets the given condition. Use probability notation to describe your result. The Z-score is at least 1.12. Click here to view page 1 of the standard normal distribution table. Click here to view page 2 of the standard normal distribution table. The probability is P(Z≥1.12) = 8686 (Round to four decimal places as needed.) 1 K In a certain region, women's heights are approximately normally distributed with mean 63.8 inches and standard deviation 4.1 inches. Men's heights are approximately normally distributed with mean 69.5 inches and standard deviation 4.6 inches. Complete parts (a) through (d) below. Click here to view page 1 of the standard normal distribution table. Click here to view page 2 of the standard normal distribution table. a. At what doorway height would 95% of men be able to pass through without having to duck? www 78.7 inches (Round to the nearest whole number as needed.) b. What percentage of women would have to duck to pass through a doorway at the height found in part (a)? .03% (Round to two decimal places as needed.) c. The height of a doorway is typically 80 inches. What percentage of men can pass through such a doorway without ducking their head? 28% (Round to two decimal places as needed.) d. The mean height of men and the mean height of women have both increased over time. For each of parts (a), (b), and (c), determine whether the correct answer to the question in the future will be larger, smaller, or impossible to know from the given information. The answer to part (a) would the answer to part (b) would increase, decrease, and the answer to part (c) would be impossible to determine. K Find the probability that a z-score randomly selected from the standard normal distribution meets the given condition. The Z-score is less than 0.85. Click to view page 1 of the table. Click to view page 2 of the table. The probability is .8023 (Round to four decimal places as needed.) K The lengths of songs played on one day on a radio station were analyzed. The lengths of the songs are approximately normally distributed with mean 249.21 seconds and standar deviation 44.54 seconds. Find the z-score of a song that is 280 seconds long. What does it mean in this situation? CI The Z-score is 69. This means that the song is 31 standard deviations longer than the mean. (Round to two decimal places as needed.) 3 The distribution of the weights of cereal in boxes of a specific brand of cereal is approximately normally distributed with mean 13 ounces and standard deviation 0.12 ounce. Complete parts (a) and (b) below. CHILE a. Explain why it would be unwise for the cereal company to advertise that each box contains 13 ounces of cereal. OA. Only half of the boxes of cereal would have exactly 13 ounces of cereal. OB. Less than half of the boxes of cereal would have at least 13 ounces of cereal. OC. More than half of the boxes of cereal would have at least 13 ounces of cereal. O D. Only half of the boxes of cereal would have at least 13 ounces of cereal. b. What should the company advertise the weight of the cereal in each box to be so that 99% of the boxes contain at least that much cereal? The company should advertise the weight of the cereal in each box to be 12.6 ounces. (Round to one decimal place as needed.) Time Remaining: 00:05:44 Next For students who graduated from high school in 2013, their scores on a standardized math test are approximately normally distributed with a mean of 505 points and a standard = deviation of 121 points. Complete parts (a) through (c) below. OA. Yes. The student's score is unusual. The associated z-score is OB. Yes. The student's score is unusual. The associated z-score is OC. No. The student's score is not unusual. The associated z-score is OD. No. The student's score is not unusual. The associated z-score is THR which is greater than the positive z-score required to determine whether an observation is unusual. which is less than the positive z-score required to determine whether an observation is unusual. .96 which is less than the positive z-score required to determine whether an observation is unusual. which is greater than the positive z-score required to determine whether an observation is unusual. b. A student who graduated from high school in 2013 scored 787 points on a standardized math test. Is that an unusual score? Explain. Select the correct choice below and fill in the answer box to complete your choice. (Round to two decimal places as needed.) OA. Yes. The student's score is unusual. The associated z-score is OB. No. The student's score is not unusual. The associated z-score is OC. No. The student's score is not unusual. The associated z-score is OD. Yes. The student's score is unusual. The associated z-score is 99 which is greater than the positive z-score required to determine whether an observation is unusual. which is less than the positive z-score required to determine whether an observation is unusual. which is greater than the positive z-score required to determine whether an observation is unusual. 99 which is less than the positive z-score required to determine whether an observation is unusual. c. For students who graduated from high school in 2013, find the cutoff for unusually high scores on a standardized math test. cording 21.04 PM Condo ead.xlsx cording 23.58 PM (a) throu (c) below. UB. Yes. The student's score is unusual. The associated z-score is OC. No. The student's score is not unusual. The associated z-score is OD. No. The student's score is not unusual. The associated z-score is TE which is less than the positive z-score required to determine whether an observation is unusual. 96 which is less than the positive z-score required to determine whether an observation is unusual. which is greater than the positive z-score required to determine whether an observation is unusual. b. A student who graduated from high school in 2013 scored 787 points on a standardized math test. Is that an unusual score? Explain. Select the correct choice below and fill in the answer box to complete your choice. (Round to two decimal places as needed.) OA. Yes. The student's score is unusual. The associated z-score is OB. No. The student's score is not unusual. The associated z-score is OC. No. The student's score is not unusual. The associated z-score is which is less than the positive z-score required to determine whether an observation is unusual. which is greater than the positive z-score required to determine whether an observation is unusual. 99 which is less than the positive z-score required to determine whether an observation is unusual. D. Yes. The student's score is unusual. The associated z-score is .99 which is greater than the positive z-score required to determine whether an observation is unusual. c. For students who graduated from high school in 2013, find the cutoff for unusually high scores on a standardized math test. 747 points (Round up to the nearest whole number.) corc Condo ead.xls cording 23.58 PM N K Without calculating P(Z < 1.3), sketch the standard normal curve to help you explain why P(Z < 1.3) cannot be equal to 0.3. Sketch the standard normal curve with the desired area shaded. Choose the correct graph below. O A. in 3-2-1 0 1 2 3 Density O Q OB. Density 0- -3-2-1 0 1 2 3 Q Q G Why can P(Z <1.3) not be equal to 0.3? OA. The area of the shaded region is greater than 0.5, so it cannot be 0.3. OB. The area of the shaded region is greater than 1, so it cannot be 0.3. OC. The area of the shaded region is less than 0.2, so it cannot be 0.3. OD. The area of the shaded region is less than 0.3, so it cannot be 0.3. Density C. 0- -3-2-1 0 1 Q M Density D. -3-2-1 0 1 2 3 Skry 6 77 K Find the probability that a z-score randomly selected from the standard normal distribution meets the given condition. Use probability notation to describe your result. The Z-score is at least 1.12. Click here to view page 1 of the standard normal distribution table. Click here to view page 2 of the standard normal distribution table. The probability is P(Z≥1.12) = 8686 (Round to four decimal places as needed.) 1 K In a certain region, women's heights are approximately normally distributed with mean 63.8 inches and standard deviation 4.1 inches. Men's heights are approximately normally distributed with mean 69.5 inches and standard deviation 4.6 inches. Complete parts (a) through (d) below. Click here to view page 1 of the standard normal distribution table. Click here to view page 2 of the standard normal distribution table. a. At what doorway height would 95% of men be able to pass through without having to duck? www 78.7 inches (Round to the nearest whole number as needed.) b. What percentage of women would have to duck to pass through a doorway at the height found in part (a)? .03% (Round to two decimal places as needed.) c. The height of a doorway is typically 80 inches. What percentage of men can pass through such a doorway without ducking their head? 28% (Round to two decimal places as needed.) d. The mean height of men and the mean height of women have both increased over time. For each of parts (a), (b), and (c), determine whether the correct answer to the question in the future will be larger, smaller, or impossible to know from the given information. The answer to part (a) would the answer to part (b) would increase, decrease, and the answer to part (c) would be impossible to determine. K Find the probability that a z-score randomly selected from the standard normal distribution meets the given condition. The Z-score is less than 0.85. Click to view page 1 of the table. Click to view page 2 of the table. The probability is .8023 (Round to four decimal places as needed.) K The lengths of songs played on one day on a radio station were analyzed. The lengths of the songs are approximately normally distributed with mean 249.21 seconds and standar deviation 44.54 seconds. Find the z-score of a song that is 280 seconds long. What does it mean in this situation? CI The Z-score is 69. This means that the song is 31 standard deviations longer than the mean. (Round to two decimal places as needed.) 3 The distribution of the weights of cereal in boxes of a specific brand of cereal is approximately normally distributed with mean 13 ounces and standard deviation 0.12 ounce. Complete parts (a) and (b) below. CHILE a. Explain why it would be unwise for the cereal company to advertise that each box contains 13 ounces of cereal. OA. Only half of the boxes of cereal would have exactly 13 ounces of cereal. OB. Less than half of the boxes of cereal would have at least 13 ounces of cereal. OC. More than half of the boxes of cereal would have at least 13 ounces of cereal. O D. Only half of the boxes of cereal would have at least 13 ounces of cereal. b. What should the company advertise the weight of the cereal in each box to be so that 99% of the boxes contain at least that much cereal? The company should advertise the weight of the cereal in each box to be 12.6 ounces. (Round to one decimal place as needed.) Time Remaining: 00:05:44 Next For students who graduated from high school in 2013, their scores on a standardized math test are approximately normally distributed with a mean of 505 points and a standard = deviation of 121 points. Complete parts (a) through (c) below. OA. Yes. The student's score is unusual. The associated z-score is OB. Yes. The student's score is unusual. The associated z-score is OC. No. The student's score is not unusual. The associated z-score is OD. No. The student's score is not unusual. The associated z-score is THR which is greater than the positive z-score required to determine whether an observation is unusual. which is less than the positive z-score required to determine whether an observation is unusual. .96 which is less than the positive z-score required to determine whether an observation is unusual. which is greater than the positive z-score required to determine whether an observation is unusual. b. A student who graduated from high school in 2013 scored 787 points on a standardized math test. Is that an unusual score? Explain. Select the correct choice below and fill in the answer box to complete your choice. (Round to two decimal places as needed.) OA. Yes. The student's score is unusual. The associated z-score is OB. No. The student's score is not unusual. The associated z-score is OC. No. The student's score is not unusual. The associated z-score is OD. Yes. The student's score is unusual. The associated z-score is 99 which is greater than the positive z-score required to determine whether an observation is unusual. which is less than the positive z-score required to determine whether an observation is unusual. which is greater than the positive z-score required to determine whether an observation is unusual. 99 which is less than the positive z-score required to determine whether an observation is unusual. c. For students who graduated from high school in 2013, find the cutoff for unusually high scores on a standardized math test. cording 21.04 PM Condo ead.xlsx cording 23.58 PM (a) throu (c) below. UB. Yes. The student's score is unusual. The associated z-score is OC. No. The student's score is not unusual. The associated z-score is OD. No. The student's score is not unusual. The associated z-score is TE which is less than the positive z-score required to determine whether an observation is unusual. 96 which is less than the positive z-score required to determine whether an observation is unusual. which is greater than the positive z-score required to determine whether an observation is unusual. b. A student who graduated from high school in 2013 scored 787 points on a standardized math test. Is that an unusual score? Explain. Select the correct choice below and fill in the answer box to complete your choice. (Round to two decimal places as needed.) OA. Yes. The student's score is unusual. The associated z-score is OB. No. The student's score is not unusual. The associated z-score is OC. No. The student's score is not unusual. The associated z-score is which is less than the positive z-score required to determine whether an observation is unusual. which is greater than the positive z-score required to determine whether an observation is unusual. 99 which is less than the positive z-score required to determine whether an observation is unusual. D. Yes. The student's score is unusual. The associated z-score is .99 which is greater than the positive z-score required to determine whether an observation is unusual. c. For students who graduated from high school in 2013, find the cutoff for unusually high scores on a standardized math test. 747 points (Round up to the nearest whole number.) corc Condo ead.xls cording 23.58 PM N K Without calculating P(Z < 1.3), sketch the standard normal curve to help you explain why P(Z < 1.3) cannot be equal to 0.3. Sketch the standard normal curve with the desired area shaded. Choose the correct graph below. O A. in 3-2-1 0 1 2 3 Density O Q OB. Density 0- -3-2-1 0 1 2 3 Q Q G Why can P(Z <1.3) not be equal to 0.3? OA. The area of the shaded region is greater than 0.5, so it cannot be 0.3. OB. The area of the shaded region is greater than 1, so it cannot be 0.3. OC. The area of the shaded region is less than 0.2, so it cannot be 0.3. OD. The area of the shaded region is less than 0.3, so it cannot be 0.3. Density C. 0- -3-2-1 0 1 Q M Density D. -3-2-1 0 1 2 3 Skry 6 77 K Find the probability that a z-score randomly selected from the standard normal distribution meets the given condition. Use probability notation to describe your result. The Z-score is at least 1.12. Click here to view page 1 of the standard normal distribution table. Click here to view page 2 of the standard normal distribution table. The probability is P(Z≥1.12) = 8686 (Round to four decimal places as needed.) 1 K In a certain region, women's heights are approximately normally distributed with mean 63.8 inches and standard deviation 4.1 inches. Men's heights are approximately normally distributed with mean 69.5 inches and standard deviation 4.6 inches. Complete parts (a) through (d) below. Click here to view page 1 of the standard normal distribution table. Click here to view page 2 of the standard normal distribution table. a. At what doorway height would 95% of men be able to pass through without having to duck? www 78.7 inches (Round to the nearest whole number as needed.) b. What percentage of women would have to duck to pass through a doorway at the height found in part (a)? .03% (Round to two decimal places as needed.) c. The height of a doorway is typically 80 inches. What percentage of men can pass through such a doorway without ducking their head? 28% (Round to two decimal places as needed.) d. The mean height of men and the mean height of women have both increased over time. For each of parts (a), (b), and (c), determine whether the correct answer to the question in the future will be larger, smaller, or impossible to know from the given information. The answer to part (a) would the answer to part (b) would increase, decrease, and the answer to part (c) would be impossible to determine. K Find the probability that a z-score randomly selected from the standard normal distribution meets the given condition. The Z-score is less than 0.85. Click to view page 1 of the table. Click to view page 2 of the table. The probability is .8023 (Round to four decimal places as needed.) K The lengths of songs played on one day on a radio station were analyzed. The lengths of the songs are approximately normally distributed with mean 249.21 seconds and standar deviation 44.54 seconds. Find the z-score of a song that is 280 seconds long. What does it mean in this situation? CI The Z-score is 69. This means that the song is 31 standard deviations longer than the mean. (Round to two decimal places as needed.) 3 The distribution of the weights of cereal in boxes of a specific brand of cereal is approximately normally distributed with mean 13 ounces and standard deviation 0.12 ounce. Complete parts (a) and (b) below. CHILE a. Explain why it would be unwise for the cereal company to advertise that each box contains 13 ounces of cereal. OA. Only half of the boxes of cereal would have exactly 13 ounces of cereal. OB. Less than half of the boxes of cereal would have at least 13 ounces of cereal. OC. More than half of the boxes of cereal would have at least 13 ounces of cereal. O D. Only half of the boxes of cereal would have at least 13 ounces of cereal. b. What should the company advertise the weight of the cereal in each box to be so that 99% of the boxes contain at least that much cereal? The company should advertise the weight of the cereal in each box to be 12.6 ounces. (Round to one decimal place as needed.) Time Remaining: 00:05:44 Next For students who graduated from high school in 2013, their scores on a standardized math test are approximately normally distributed with a mean of 505 points and a standard = deviation of 121 points. Complete parts (a) through (c) below. OA. Yes. The student's score is unusual. The associated z-score is OB. Yes. The student's score is unusual. The associated z-score is OC. No. The student's score is not unusual. The associated z-score is OD. No. The student's score is not unusual. The associated z-score is THR which is greater than the positive z-score required to determine whether an observation is unusual. which is less than the positive z-score required to determine whether an observation is unusual. .96 which is less than the positive z-score required to determine whether an observation is unusual. which is greater than the positive z-score required to determine whether an observation is unusual. b. A student who graduated from high school in 2013 scored 787 points on a standardized math test. Is that an unusual score? Explain. Select the correct choice below and fill in the answer box to complete your choice. (Round to two decimal places as needed.) OA. Yes. The student's score is unusual. The associated z-score is OB. No. The student's score is not unusual. The associated z-score is OC. No. The student's score is not unusual. The associated z-score is OD. Yes. The student's score is unusual. The associated z-score is 99 which is greater than the positive z-score required to determine whether an observation is unusual. which is less than the positive z-score required to determine whether an observation is unusual. which is greater than the positive z-score required to determine whether an observation is unusual. 99 which is less than the positive z-score required to determine whether an observation is unusual. c. For students who graduated from high school in 2013, find the cutoff for unusually high scores on a standardized math test. cording 21.04 PM Condo ead.xlsx cording 23.58 PM (a) throu (c) below. UB. Yes. The student's score is unusual. The associated z-score is OC. No. The student's score is not unusual. The associated z-score is OD. No. The student's score is not unusual. The associated z-score is TE which is less than the positive z-score required to determine whether an observation is unusual. 96 which is less than the positive z-score required to determine whether an observation is unusual. which is greater than the positive z-score required to determine whether an observation is unusual. b. A student who graduated from high school in 2013 scored 787 points on a standardized math test. Is that an unusual score? Explain. Select the correct choice below and fill in the answer box to complete your choice. (Round to two decimal places as needed.) OA. Yes. The student's score is unusual. The associated z-score is OB. No. The student's score is not unusual. The associated z-score is OC. No. The student's score is not unusual. The associated z-score is which is less than the positive z-score required to determine whether an observation is unusual. which is greater than the positive z-score required to determine whether an observation is unusual. 99 which is less than the positive z-score required to determine whether an observation is unusual. D. Yes. The student's score is unusual. The associated z-score is .99 which is greater than the positive z-score required to determine whether an observation is unusual. c. For students who graduated from high school in 2013, find the cutoff for unusually high scores on a standardized math test. 747 points (Round up to the nearest whole number.) corc Condo ead.xls cording 23.58 PM N K Without calculating P(Z < 1.3), sketch the standard normal curve to help you explain why P(Z < 1.3) cannot be equal to 0.3. Sketch the standard normal curve with the desired area shaded. Choose the correct graph below. O A. in 3-2-1 0 1 2 3 Density O Q OB. Density 0- -3-2-1 0 1 2 3 Q Q G Why can P(Z <1.3) not be equal to 0.3? OA. The area of the shaded region is greater than 0.5, so it cannot be 0.3. OB. The area of the shaded region is greater than 1, so it cannot be 0.3. OC. The area of the shaded region is less than 0.2, so it cannot be 0.3. OD. The area of the shaded region is less than 0.3, so it cannot be 0.3. Density C. 0- -3-2-1 0 1 Q M Density D. -3-2-1 0 1 2 3 Skry 6 77 K Find the probability that a z-score randomly selected from the standard normal distribution meets the given condition. Use probability notation to describe your result. The Z-score is at least 1.12. Click here to view page 1 of the standard normal distribution table. Click here to view page 2 of the standard normal distribution table. The probability is P(Z≥1.12) = 8686 (Round to four decimal places as needed.) 1 K In a certain region, women's heights are approximately normally distributed with mean 63.8 inches and standard deviation 4.1 inches. Men's heights are approximately normally distributed with mean 69.5 inches and standard deviation 4.6 inches. Complete parts (a) through (d) below. Click here to view page 1 of the standard normal distribution table. Click here to view page 2 of the standard normal distribution table. a. At what doorway height would 95% of men be able to pass through without having to duck? www 78.7 inches (Round to the nearest whole number as needed.) b. What percentage of women would have to duck to pass through a doorway at the height found in part (a)? .03% (Round to two decimal places as needed.) c. The height of a doorway is typically 80 inches. What percentage of men can pass through such a doorway without ducking their head? 28% (Round to two decimal places as needed.) d. The mean height of men and the mean height of women have both increased over time. For each of parts (a), (b), and (c), determine whether the correct answer to the question in the future will be larger, smaller, or impossible to know from the given information. The answer to part (a) would the answer to part (b) would increase, decrease, and the answer to part (c) would be impossible to determine. K Find the probability that a z-score randomly selected from the standard normal distribution meets the given condition. The Z-score is less than 0.85. Click to view page 1 of the table. Click to view page 2 of the table. The probability is .8023 (Round to four decimal places as needed.) K The lengths of songs played on one day on a radio station were analyzed. The lengths of the songs are approximately normally distributed with mean 249.21 seconds and standar deviation 44.54 seconds. Find the z-score of a song that is 280 seconds long. What does it mean in this situation? CI The Z-score is 69. This means that the song is 31 standard deviations longer than the mean. (Round to two decimal places as needed.)

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Part 1 a Since for normal distribution meanmedian only half of the boxes of cereal would have at lea... View the full answer