Question
1. Suppose the average number of returns processed by employees of a tax preparation service during tax season is 11 per day with a standard
1. Suppose the average number of returns processed by employees of a tax preparation service during tax season is
11
per day with a standard deviation of
5
per day. A random sample of 36 employees taken during tax season revealed the number of returns processed daily shown below. Use these data to answer parts a through c.
10 | 13 | 18 | 14 | 10 | 11 | 12 | 10 | 18 | 13 | 14 | 16 | 11 | 11 | 16 | 13 | 15 | 13 | |
10 | 9 | 8 | 16 | 13 | 11 | 9 | 12 | 13 | 10 | 10 | 9 | 6 | 4 | 13 | 8 | 10 | 5 |
a. What is the probability of having a sample mean equal to or smaller than the sample mean for this sample if the population mean is
11
processed returns daily with a standard deviation of
5
returns per day?
The probability is
enter your response here.
(Round to four decimal places as needed.)
b. What is the probability of having a sample mean larger than the sample mean for this sample if the population mean is
11
processed returns daily with a standard deviation of
5
returns per day?
The probability is
enter your response here.
(Round to four decimal places as needed.)
c. Explain how it is possible to answer parts a and b when the population distribution of daily tax returns at the tax firm is not known. Choose the correct answer below.
A.
All data are normally distributed. Thus, the distribution of the sample means will be approximately normal.
B.
The Central Limit Theorem can be applied because the sample size is sufficiently large. Thus, the distribution of the sample means will be either skewed to the left or skewed to the right.
C.
The Central Limit Theorem can be applied because the sample size is not large. Thus, the distribution of the sample means will be approximately normal.
D.
The Central Limit Theorem can be applied because the sample size is sufficiently large. Thus, the distribution of the sample means will be approximately normal.
2.Suppose nine items are sampled from a normally distributed population with a mean of
96
and a standard deviation of
15.
The nine randomly sampled values are shown in the table.
112 | 104 | 75 | 89 | 92 | |
99 | 88 | 52 | 67 |
Calculate the probability of getting a sample mean that is smaller than the mean for these nine sampled values.
The probability that a sample mean is smaller than the mean for these nine values is
(Round to four decimal places as needed.)
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