Final answer:
Using the Central Limit Theorem, we know the sample mean for the waiting time will approximate a normal distribution with a mean of 90 seconds and a standard deviation of 9 seconds. We can calculate the probability of observing an average waiting time of 110 seconds or more under this distribution to assess the accuracy of the cashier's claim.
Step-by-step explanation:
The Central Limit Theorem (CLT) states that the distribution of sample means approximates a normal distribution as the sample size gets larger, regardless of the shape of the population distribution. Since the waiting time in the store has an exponential distribution with mean μ = 1/Λ and variance σ2 = 1/Λ2, and we have a large sample size (n = 100), we can use the CLT to approximate the distribution of the sample mean.
(a) According to CLT, the distribution of the average waiting time for 100 customers will be normally distributed with a mean of 90 seconds and standard deviation σ/√n = (90 seconds)/√100 = 9 seconds.
(b) Assuming the cashier's claim is accurate, the probability that the average waiting time for a sample of 100 customers is greater than or equal to 110 seconds can be found using the standard normal distribution (Z-score).
Since our sample mean (110 seconds) is greater than the claimed mean (90 seconds), we need to calculate the Z-score and find the corresponding tail probability.
(c) If the probability found in (b) is very low, it might suggest that the cashier's claim is not accurate as the observed average is unlikely given the claimed distribution.