New Semester
Started
Get
50% OFF
Study Help!
--h --m --s
Claim Now
Question Answers
Textbooks
Find textbooks, questions and answers
Oops, something went wrong!
Change your search query and then try again
S
Books
FREE
Study Help
Expert Questions
Accounting
General Management
Mathematics
Finance
Organizational Behaviour
Law
Physics
Operating System
Management Leadership
Sociology
Programming
Marketing
Database
Computer Network
Economics
Textbooks Solutions
Accounting
Managerial Accounting
Management Leadership
Cost Accounting
Statistics
Business Law
Corporate Finance
Finance
Economics
Auditing
Tutors
Online Tutors
Find a Tutor
Hire a Tutor
Become a Tutor
AI Tutor
AI Study Planner
NEW
Sell Books
Search
Search
Sign In
Register
study help
statistics
elementary statistics a step by step approach
Elementary Statistics A Step By Step Approach 9th Edition Allan G. Bluman - Solutions
Find the values for each.a. ta/2 and n = 18 for the 99% confidence interval for the meanb. ta/2 and n = 23 for the 95% confidence interval for the meanc. ta/2 and n = 15 for the 98% confidence interval for the meand. ta/2 and n = 10 for the 90% confidence interval for the meane. ta/2 and n = 20 for
What is meant by degrees of freedom?
Use the top home run hitter from each major league baseball team as the data set. Find the mean and the standard deviation for the number of home runs hit by the top hitter on each team. Find a 95% confidence interval for the mean number of home runs hit.
Determine whether each statement is true or false. If the statement is false, explain why.For a specific confidence interval, the larger the sample size, the smaller the margin of error will be.
Find p̂ and q̂ for each percentage.a. n = 60 and X = 35b. n = 95 and X = 43c. 68%d. 55%e. 12%
Repeat Exercise 1, using a different variable and a sample of 15.Data from in Exercise 1From the Data Bank choose a variable, find the mean, and construct the 95 and 99% confidence intervals of the population mean. Use a sample of at least 30 subjects. Find the mean of the population, and determine
Use 30 stocks classified as the Dow Jones industrials as the sample. Note the amount each stock has gained or lost in the last quarter. Compute the mean and standard deviation for the data set. Compute the 95% confidence interval for the mean and the 95% confidence interval for the standard
A confidence interval for a median can be found by using these formulasto define positions in the set of ordered data values. Suppose a data set has 30 values, and you want to find the 95% confidence interval for the median. Substituting in the formulas, you getArrange the data in order from
Determine whether each statement is true or false. If the statement is false, explain why.Interval estimates are preferred over point estimates since a confidence level can be specified.
In each case, find p̂ and q̂ .a. n = 80 and X = 40b. n = 200 and X = 90c. n = 130 and X = 60d. 25%e. 42%
From the Data Bank choose a variable, find the mean, and construct the 95 and 99% confidence intervals of the population mean. Use a sample of at least 30 subjects. Find the mean of the population, and determine whether it falls within the confidence interval.
Find P(z > 2.3 and z < - 1.2).
Find P(z < 2.3 or z > - 1.2).
For Exercises z0 is the statistical notation for an unknown z value. It serves that same function as x does in an algebraic equation.Find z0 such that P( - z0 < z < z0) = 0.76.
For Exercises z0 is the statistical notation for an unknown z value. It serves that same function as x does in an algebraic equation.Find z0 such that the area between z0 and z =-0.5 is 0.2345 (two answers).
For Exercises z0 is the statistical notation for an unknown z value. It serves that same function as x does in an algebraic equation.Find z0 such that P(z0 < z < 2.5) = 0.7672.
For Exercises z0 is the statistical notation for an unknown z value. It serves that same function as x does in an algebraic equation.Find z0 such that P( - 1.2 <z <z0) = 0.8671.
For Exercises find the z value that corresponds to the given area. 0.0188- N 0
The average GMAT scores for the top-30 ranked graduate schools of business are listed here. Check for normality. 718 703 703 703 700 690 695 705 690 688 676 681 689 686 691 669 674 652 680 670 651 651 637 662 641 645 645 642 660 636
In Exercises find the probabilities for each, using the standard normal distribution.P(1.12 < z < 1.43)
The number of calories contained in a selection of fast-food sandwiches is shown here. Check for normality. 390 540 535 390 320 430 405 225 660 675 460 530 580 720 530 530 290 300 470 290 1010 340 320 560 440 450 610
In Exercises find the probabilities for each, using the standard normal distribution.P(1.51 < z < 2.17)
In Exercises find the probabilities for each, using the standard normal distribution.P(z <- 1.77)
The mean temperature (of daily maximum temperatures) in July for Dallas–Ft. Worth, Texas, is 85 degrees. Assuming a normal distribution, what would the standard deviation have to be if 10% of days have a high of at least 100 degrees?
In Exercises find the probabilities for each, using the standard normal distribution.P(z <- 1.46)
In Exercises find the probabilities for each, using the standard normal distribution.P(z > 0.82)
Fifty-three percent of U.S. households have a personal computer. In a random sample of 250 households, what is the probability that fewer than 120 have a PC?
In Exercises find the probabilities for each, using the standard normal distribution.P(z > 2.51)
The percentage of U.S. households that have online connections is 44.9%. In a random sample of 420 households, what is the probability that fewer than 200 have online connections?
If 8% of all people in a certain geographic region are unemployed, find the probability that in a sample of 200 people, fewer than 10 people are unemployed.
In Exercises find the probabilities for each, using the standard normal distribution.P( - 1.43 < z < 0)
The average electric bill in a residential area is $72 for the month of April. The standard deviation is $6. If the amounts of the electric bills are normally distributed, find the probability that the mean of the bill for 15 residents will be less than $75.
In Exercises find the probabilities for each, using the standard normal distribution.P(0 < z < 0.92)
For Exercises find the area under the standard normal distribution curve.To the left of z = - 2.15 and to the right of z = 1.62
Membership in an elite organization requires a test score in the upper 30% range. If μ = 115 and σ = 12, find the lowest acceptable score that would enable a candidate to apply for membership. Assume the variable is normally distributed.
The average thickness of books on a library shelf is 8.3 centimeters. The standard deviation is 0.6 centimeter. If 20% of the books are oversized, find the minimum thickness of the oversized books on the library shelf. Assume the variable is normally distributed.
On the daily run of an express bus, the average number of passengers is 48. The standard deviation is 3. Assume the variable is normally distributed. Find the probability that the bus will havea. Between 36 and 40 passengersb. Fewer than 42 passengersc. More than 48 passengersd. Between 43 and 47
The normal distribution can be used to approximate the binomial distribution when n · p and n · q are both greater than or equal to ________ .
For Exercises find the area under the standard normal distribution curve.To the right of z = - 0.12
For Exercises find the area under the standard normal distribution curve.To the left of z = 1.31
The average number of gallons of lemonade consumed by the football team during a game is 20, with a standard deviation of 3 gallons. Assume the variable is normally distributed. When a game is played, find the probability of usinga. Between 20 and 25 gallonsb. Less than 19 gallonsc. More than 21
For Exercises find the area under the standard normal distribution curve.To the left of z = 2.22
The average height of a certain age group of people is 53 inches. The standard deviation is 4 inches. If the variable is normally distributed, find the probability that a selected individual’s height will bea. Greater than 59 inchesb. Less than 45 inchesc. Between 50 and 55 inchesd. Between 58
For Exercises find the area under the standard normal distribution curve.Between z = - 1.46 and z = - 1.98
The average amount of rain per year in Greenville is 49 inches. The standard deviation is 8 inches. Find the probability that next year Greenville will receive the following amount of rainfall. Assume the variable is normally distributed.a. At most 55 inches of rainb. At least 62 inches of rainc.
For Exercises find the area under the standard normal distribution curve.Between z = - 0.24 and z = - 1.12
Using the standard normal distribution, find each probability.a. P(0 < z < 2.16)b. P( - 1.87 < z < 0)c. P( - 1.63 < z < 2.17)d. P(1.72 < z < 1.98)e. P( - 2.17 < z < 0.71)f. P(z > 1.77)g. P(z < - 2.37)h. P(z > - 1.73)i. P(z
For Exercises find the area under the standard normal distribution curve.Between z = 0.96 and z = 0.36
Find the area under the standard normal distribution for each.a. Between 0 and 1.50b. Between 0 and - 1.25c. Between 1.56 and 1.96d. Between - 1.20 and - 2.25e. Between - 0.06 and 0.73f. Between 1.10 and - 1.80g. To the right of z = 1.75h. To the right of z = - 1.28i. To the left of z = - 2.12j. To
For Exercises find the area under the standard normal distribution curve.Between z = 1.56 and z = 1.83
The average annual professor’s salary at a doctoral level at a private, independent institution is $159,964 for men and $147,702 for women. Consider the women’s salaries. Assume that they are normally distributed with a standard deviation of $8900. What is the probability that a woman professor
The correction factor for the central limit theorem should be used when the sample size is greater than ________ of the size of the population.
For Exercises find the area under the standard normal distribution curve.Between z = 1.23 and z = 1.90
For Exercises find the area under the standard normal distribution curve.Between z = 1.09 and z = 1.83
The standard deviation of all possible sample means is called the ________.
According to recent surveys, 60% of households have personal computers. If a random sample of 180 households is selected, what is the probability that more than 60 but fewer than 100 have a personal computer?
When a distribution is positively skewed, the relationship of the mean, median, and mode from left to right will bea. Mean, median, modeb. Mode, median, mean c. Median, mode, meand. Mean, mode, median
For Exercises find the area under the standard normal distribution curve.To the left of z = 0.75
The mean of the sample means equals ________.
In a recent year, 56% of employers offered a consumer-directed health plan (CDHP). This type of plan typically combines a high deductible with a health savings plan. Choose 80 employers at random. What is the probability that more than 50 will offer a CDHP?
In a recent year the average movie ticket cost $7.89. In a random sample of 50 movie tickets from various areas, what is the probability that the mean cost exceeds $8.00, given that the population standard deviation is $1.39?For Exercises assume that the sample is taken from a large population and
For Exercises find the area under the standard normal distribution curve.To the left of z = 1.39
Each American uses an average of 650 pounds (295 kg) of paper in a year. Suppose that the distribution is approximately normal with a population standard deviation of 153.5 pounds. Assume the variable is normally distributed. Find the probability that a randomly selected American usesa. More than
The difference between a sample mean and a population mean is due to ________.
For Exercises find the area under the standard normal distribution curve.To the right of z = 2.01
When one is using the standard normal distribution, P(z < 0) = ________.
For Exercises find the area under the standard normal distribution curve.To the right of z = 0.29
The standard deviation of all possible sample means equalsa. The population standard deviationb. The population standard deviation divided by the population meanc. The population standard deviation divided by the square root of the sample sized. The square root of the population standard deviation
The mean home price in Raleigh, North Carolina, is $217,600. Assuming that the home prices are normally distributed with a standard deviation of $36,400, what is the probability that a randomly selected home in Raleigh has a price below $200,000? Below $150,000?
For Exercises find the area under the standard normal distribution curve.Between z = 0 and z = 0.32
In a recent year, 23.3% of Americans smoked cigarettes. What is the probability that in a random sample of 200 Americans, more than 50 smoke?
Use the normal approximation to find the probabilities for the specific value(s) of X.a. n = 10, p = 0.5, X ≥ 7b. n = 20, p = 0.7, X ≤ 12c. n = 50, p = 0.6, X ≤ 40
For Exercises find the area under the standard normal distribution curve.Between z = 0 and z = 2.14
Which is not a property of the standard normal distribution?a. It’s symmetric about the mean.b. It’s uniform.c. It’s bell-shaped.d. It’s unimodal.
Explain why the method used in step 7 works.
For Exercises find the area under the standard normal distribution curve.Between z = 0 and z = 1.77
Approximately what percentage of normally distributed data values will fall within 1 standard deviation above or below the mean?a. 68%b. 95% c. 99.7%d. Variable
To find an approximation of the standard deviation, locate the values on the x axis that correspond to the 16 and 84% values on the y axis. Subtract these two values and divide the result by 2. Compare this approximate standard deviation to the computed standard deviation.
The average weekly unemployment benefit in Montana is $272. Suppose that the benefits are normally distributed with a standard deviation of $43. A random sample of 15 benefits is chosen in Montana. What is the probability that the mean for this sample is greater than the U.S. average, which is
For Exercises find the area under the standard normal distribution curve.Between z = 0 and z = 0.98
The mean of the standard normal distribution isa. 0b. 1 c. 100d. Variable
Confirm the two formulas hold true for the central limit theorem for the population containing the elements {1, 5, 10}. First, compute the population mean and standard deviation for the data set. Next, create a list of all 9 of the possible two-element samples that can be created with replacement:
To find an approximation of the mean or median, draw a horizontal line from the 50% point on the y axis over to the curve and then a vertical line down to the x axis. Compare this approximation of the mean with the computed mean.
Collect data regarding Math SAT scores to complete this problem. What are the mean and standard deviation for statewide Math SAT scores? What SAT score separates the bottom 10% of states from the others? What is the probability that a randomly selected state has a statewide SAT score above 500?
Determine whether each statement is true or false. If the statement is false, explain why.The central limit theorem applies to means of samples selected from different populations.
If the points fall approximately in a straight line, it can be concluded that the distribution is normal. Do you feel that this distribution is approximately normal? Explain your answer.
Determine whether each statement is true or false. If the statement is false, explain why.The area under the standard normal distribution to the left of z = 0 is negative.
When is the normal distribution not a good approximation for the binomial distribution?
Use the data regarding heart rates collected in data project 6 of Chapter 2 for this problem. Use the sample mean and standard deviation as estimates of the population parameters. For the before-exercise data, what heart rate separates the top10% from the other values? For the after-exercise data,
Using the normal probability paper shown in Table 6–3, label the x axis with the class boundaries as shown and plot the percents.Data from in Table 6-3 TABLE 6-3 98 99 96 90 80 70 30 40 50 60 Oz 10 5 2 1 Normal Probability Paper 89.5 104.5 119.5 134.5 149.5 164.5 179.5
Determine whether each statement is true or false. If the statement is false, explain why.The z value corresponding to a number below the mean is always negative.
Using the standard normal distribution, find each probability.a. P(0 < z < 2.07)b. P( - 1.83 < z < 0)c. P( - 1.59 < z < 2.01)d. P(1.33 < z < 1.88)e. P( - 2.56 < z < 0.37)
Use the data collected in data project 3 of Chapter 2 regarding song lengths. If the sample estimates for mean and standard deviation are used as replacements for the population parameters for this data set, what song length separates the bottom 5% and top 5% from the other values?Data from in
Find the cumulative percents for each class by dividing each cumulative frequency by 200 (the total frequencies) and multiplying by 100%. (For the first class, it would be 24/200 ×100% = 12%.) Place these values in the last column.
What is the mean of the sample means?
Determine whether each statement is true or false. If the statement is false, explain why.All variables that are approximately normally distributed can be transformed to standard normal variables.
Find the mean and standard deviation for the batting average for a player in the most recently completed MLB season. What batting average would separate the top 5% of all hitters from the rest? What is the probability that a randomly selected player bats over 0.300? What is the probability that a
Find the cumulative frequencies for each class, and place the results in the third column.
Determine whether each statement is true or false. If the statement is false, explain why.The standard normal distribution is a continuous distribution.
Use the data collected in data project 1 of Chapter 2 regarding earnings per share to complete this problem. Use the mean and standard deviation computed in data project 1 of Chapter 3 as estimates for the population parameters. What value separates the top 5% of stocks from the others?Data from in
Showing 1400 - 1500
of 1910
First
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
Step by Step Answers
Get 50% OFF
Claim Now
