Question: Please help me with this question. Thank you so much! Question 6 1 pts In N(5, 4): For Z-scores: type in exact answers as decimals.

Please help me with this question. Thank you so much!

Please help me with this question. Thank you so much! Question 61 pts In N(5, 4): For Z-scores: type in exact answers asdecimals. For probabilities: round to 4 decimal places. P(-1 3) = P(Z> P(X , or = HI OI OI 15 (Due to roundingmake sure to compare only to 2 decimal places).Question 4 1 ptsThe distribution of IQ scores of the population is normally distributed witha mean of 100 and a standard deviation of 15 for aparticular data value. Since the population is normally distributed all sampling distributionswill also be normally distributed. Find the z-scores for each of thefollowing. Make sure to pay attention to the distribution you should use.Write exact answers as decimals. A single person with an IQ of121: Z-score = Mean of a sample of 4 IQ's is 121:Z-score = Mean of a sample of 9 IQ's is 121: Z-score= Mean of a sample of 25 IQ's is 121: Z-score =

Question 6 1 pts In N(5, 4): For Z-scores: type in exact answers as decimals. For probabilities: round to 4 decimal places. P(-1 3) = P(Z > P(X , or = HI OI OI 15 (Due to rounding make sure to compare only to 2 decimal places).Question 4 1 pts The distribution of IQ scores of the population is normally distributed with a mean of 100 and a standard deviation of 15 for a particular data value. Since the population is normally distributed all sampling distributions will also be normally distributed. Find the z-scores for each of the following. Make sure to pay attention to the distribution you should use. Write exact answers as decimals. A single person with an IQ of 121: Z-score = Mean of a sample of 4 IQ's is 121: Z-score = Mean of a sample of 9 IQ's is 121: Z-score = Mean of a sample of 25 IQ's is 121: Z-score = Mean of a sample of 100 IQ's is 121: Z-score = Mean of a sample of 625 IQ's is 121: Z-score = A single person with an IQ of 106:2-score =Question 21 0.5 pts Suppose that the hibernation time for black bears is normally distributed with mean 150 days and standard deviation 30 days. 12 randomly selected black bears are monitored. Round all answers to two decimal places. A. X ~ N( 150 8.66 B. For the 12 black bears, find the probability that the average hibernation time is more than 170 days. 0.01 C. What is the probability that one randomly selected black bear will hibernate longer than 170 days? 0.25 D. Find the IQR for the average of 12 bears. Q1 = 144.15 Q3 = 155.85 IQR: 11.69 Some Helpful Videos: Finding the Sampling Distribution , Finding a Probability Using the Central Limit Theorem e , Finding a Sample Mean Given its Percentile Hint Textbook PagesQuestion 3 2 pts Population values: 1, 5, 6 H = Population mean= o = Population standard deviation = (Round to two places after the decimal). Make sure to enter the population standard deviation and not the sample standard deviation. Sampling distribution of size n = 3: Ordered Possible Samples size 2 Sample Mean 1, 1 1, 5 1, 6 3.5 5 5 5 5 5.5 6, 3.5\fQuestion 5 1 pts In N(5,4): Round Z-scores to 2 places after the decimal and use rounded value to find the data value in N(5,4). If P(X a) = .02, then the z-score for a = and the value in N(5,4) = If P(b , or = O OI (Due to rounding make sure to compare only to 2 decimal places).Question 10 0.5 pts If X is normally distributed with a mean of 80 and a standard deviation of 8.88. If you take a sample of size 240, what is the mean of the sampling distribution? Write your answer correct to 4 decimal places.Question 2 2 pts Population values: 2, 4, 6 A = Population mean = o = Population standard deviation = (Round to two places after the decimal). Make sure to enter the population standard deviation and not the sample standard deviation. Sampling distribution of size n = 2: Ordered Possible Samples size 2 Sample Mean 2, 2 2, 4 3 2. 4 4. 2 3 4. 4.6 6. 4

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