A normal population has a mean of 75 and a standard deviation of 5. You select a sample of 40. Compute the probability the sample mean is: a. Less than 74. b. Between 74 and 76. c. Between 76 and 77. d. Greater than 77

A set of data has a mean of 75 and a standard deviation of 5. You know nothing else about the size of the data set or the shape of the data distribution. a. What can you say about the proportion of measurements that fall between 60 and 90? b. What can you say about the proportion of measurements that fall between 65 and 85? c. What can you say about the proportion of measurements that are less than 65?

A normal population has a mean of 75 and a standard deviation of 5. You choose a 40-person sample. Calculate the likelihood that the sample mean is between 76 and 77.

A normal population has a mean of 80.0 and a standard deviation of 14.0. a. Compute the probability of a value between 75.0 and 90.0. b. Compute the probability of a value 75.0 or less. c. Compute the probability of a value between 55.0 and 70.0

A population consists of the following five values: 0, 0, 1, 3, 6. a. List all samples of size 3, and compute the mean of each sample. b. Compute the mean of the distribution of sample means and the population mean. Compare the two values. c. Compare the dispersion in the population with that of the sample means

A normal population has a mean of 75 and a standard deviation of 5. You choose a 40-person sample. Calculate the likelihood that the sample mean is less than 74.

We want to estimate the population mean within 5, with a 99 percent level of confidence. The population standard deviation is estimated to be 15. How large a sample is required?

A normal population has a mean of 20.0 and a standard deviation of 4.0. a. Compute the z value associated with 25.0. b. What proportion of the population is between 20.0 and 25.0? c. What proportion of the population is less than 18.0?

A normal population has a mean of 75 and a standard deviation of 5. You choose a 40-person sample. Calculate the likelihood that the sample mean is between 74 and 76

Suppose you know IT and you want an 85% confidence level. What value would you use as z in formula (9-1)?

Formula (9-1)- The Bun-and-Run is a franchise fast-food restaurant located in the Northeast specializing in half-pound hamburgers, fish sandwiches, and chicken sandwiches. Soft drinks and French fries are also available. The Planning Department of Bun-and-Run Inc. reports that the distribution of daily sales for restaurants follows the normal distribution and that the population standard deviation is $\$ 3,000$. A sample of 40 showed the mean daily sales to be $\$ 20,000$.

(a) What is the population mean?

(b) What is the best estimate of the population mean? What is this value called?

(c) Develop a $95 \%$ confidence interval for the population mean.

(d) Interpret the confidence interval.

There are five sales associates at Mid-Motors Ford. The five associates and the number of cars they sold last week are:

$$\begin{array}{lc} \text { Sales Associate } & \text { Cars Sold } \\ \hline \text { Peter Hankish } & 8 \\ \text { Connie Stallter } & 6 \\ \text { Juan Lopez } & 4 \\ \text { Ted Barnes } & 10 \\ \text { Peggy Chu } & 6 \end{array}$$

a. How many different samples of size 2 are possible? b. List all possible samples of size 2 , and compute the mean of each sample. c. Compare the mean of the sampling distribution of the sample mean with that of the population. d. On a chart similar to Chart $8-2$, compare the dispersion in sample means with that of the population.

In a binomial distribution, n = 8 and π = .30. Find the probabilities of the following events. a. x =2. b. x =2 c. x =3.

In a binomial distribution, n = 12 and π = .60. Find the following probabilities. a. x =5. b. x =5. c. x =6

Recent studies indicate that the typical 50-year-old woman spends $350 per year for personal-care products. The distribution of the amounts spent follows a normal distribution with a standard deviation of$45 per year. We select a random sample of 40 women. The mean amount spent for those sampled is $335. What is the likelihood of finding a sample mean this large or larger from the specified population?