The average amount of money spent on a dinner in Fresno is $\$ 17.00$ with a population standard deviation of $\$ 4.00$. You randomly select 100 diners.
a. What is the probability that one of the diners spent between $\$ 16.32$ and $\$ 18.24$ on his/her dinner?
b. What is the probability that the average amount the 100 diners spent on dinner was between $\$ 16.32$ and $\$ 18.24$ ?
Answer
b. The probability that the average amount the 100 diners spent on dinner was between \$16.32 and \$18.24 is approximately \(\boxed{0.954}\).
Step 1 ：Given that the average amount of money spent on a dinner in Fresno is \$17.00 with a population standard deviation of \$4.00. We randomly select 100 diners.
Step 2 ：For part a, we need to calculate the z-scores for \$16.32 and \$18.24. The z-score is calculated as \((X - \mu) / \sigma\), where X is the value, \(\mu\) is the mean, and \(\sigma\) is the standard deviation.
Step 3 ：Substituting the given values, we get z-scores as -0.17 and 0.31 respectively.
Step 4 ：Using the standard normal distribution table, we find the probability that a z-score falls between these two values is approximately 0.189.
Step 5 ：For part b, we need to calculate the z-scores for \$16.32 and \$18.24. However, because we are dealing with a sample mean, the standard deviation is divided by the square root of the sample size.
Step 6 ：Substituting the given values, we get z-scores as -1.70 and 3.10 respectively.
Step 7 ：Using the standard normal distribution table, we find the probability that a z-score falls between these two values is approximately 0.954.
Step 8 ：Final Answer: a. The probability that one of the diners spent between \$16.32 and \$18.24 on his/her dinner is approximately \(\boxed{0.189}\).
Step 9 ：b. The probability that the average amount the 100 diners spent on dinner was between \$16.32 and \$18.24 is approximately \(\boxed{0.954}\).
