of last try: 3.6 of 6 pts. See Details for more. Next question $\rightleftarrows$ Get a similar question You can retry this question below he average price of a college math textbook is $\$ 161$ and the standard deviation is $\$ 30$. Suppose that 18 extbooks are randomly chosen. Round all answers to 4 decimal places where possible. a. What is the distribution of $\bar{x} ? \bar{x}-N\left(161 \quad \checkmark, 7.0711\right.$ ) of $o^{s}$ b. For the group of 18 , find the probability that the average price is between $\$ 160$ and $\$ 171$. c. Find the third quartile for the average textbook price for this sample size. $\$$ nearest cent) d. For part b), is the assumption that the distribution is normal necessary? (round to the

Step 1 ：First, we need to understand the distribution of the sample mean. The sample mean follows a normal distribution with mean equal to the population mean and standard deviation equal to the population standard deviation divided by the square root of the sample size. In this case, the population mean is \$161 and the population standard deviation is \$30. The sample size is 18.

Step 2 ：Next, we calculate the standard deviation of the sample mean. This is given by \(\frac{\sigma}{\sqrt{n}}\), where \(\sigma\) is the population standard deviation and \(n\) is the sample size. Substituting the given values, we get \(\frac{30}{\sqrt{18}}\approx 7.0711\).

Step 3 ：So, the distribution of the sample mean is \(N(161, 7.0711)\).

Step 4 ：For part b, we need to find the probability that the sample mean is between \$160 and \$171. To do this, we first convert these values to z-scores. The z-score is given by \(\frac{x-\mu}{\sigma}\), where \(x\) is the value, \(\mu\) is the mean, and \(\sigma\) is the standard deviation.

Step 5 ：The z-score for \$160 is \(\frac{160-161}{7.0711}\approx -0.1414\) and the z-score for \$171 is \(\frac{171-161}{7.0711}\approx 1.4142\).

Step 6 ：We then look up these z-scores in the standard normal distribution table to find the corresponding probabilities. The probability for a z-score of -0.1414 is approximately 0.4443 and the probability for a z-score of 1.4142 is approximately 0.9213.

Step 7 ：The probability that the sample mean is between \$160 and \$171 is the difference between these two probabilities, which is \(0.9213 - 0.4443 = 0.4770\).

Step 8 ：For part c, the third quartile is the value below which 75% of the data fall. In a standard normal distribution, this corresponds to a z-score of approximately 0.6745. We convert this z-score back to a value in the original distribution using the formula \(x = \mu + z\sigma\). Substituting the given values, we get \(x = 161 + 0.6745*7.0711\approx 166.77\). So, the third quartile for the average textbook price for this sample size is approximately \$166.77.

Step 9 ：For part d, the assumption that the distribution is normal is necessary because we used z-scores and the standard normal distribution table to calculate probabilities. These tools are based on the properties of the normal distribution. If the distribution is not normal, these calculations may not be accurate.

Step 10 ：\(\boxed{\text{a. }\bar{x} \sim N(161, 7.0711)}\)

Step 11 ：\(\boxed{\text{b. }P(160 \leq \bar{x} \leq 171) = 0.4770}\)

Step 12 ：\(\boxed{\text{c. }Q3 = \$166.77}\)

Step 13 ：\(\boxed{\text{d. }\text{Yes, the assumption that the distribution is normal is necessary.}}\)