Answer:
1.32% probability that a randomly selected 100 customers will spend an average of less than $50 on dinner
Step-by-step explanation:
To solve this question, we need to understand the normal probability distribution and the central limit theorem.
Normal probability distribution:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean
and standard deviation
, the zscore of a measure X is given by:
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
Central limit theorem
The Central Limit Theorem estabilishes that, for a random variable X, with mean
and standard deviation
, a large sample size can be approximated to a normal distribution with mean
and standard deviation ![s = \frac{\sigma}{\sqrt{n}}](https://tex.z-dn.net/?f=s%20%3D%20%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D)
In this problem, we have that:
![\mu = 54, \sigma = 18, n = 100, s = \frac{18}{\sqrt{100}} = 1.8](https://tex.z-dn.net/?f=%5Cmu%20%3D%2054%2C%20%5Csigma%20%3D%2018%2C%20n%20%3D%20100%2C%20s%20%3D%20%5Cfrac%7B18%7D%7B%5Csqrt%7B100%7D%7D%20%3D%201.8)
Based on this information, what is the probability that a randomly selected 100 customers will spend an average of less than $50 on dinner?
This probability is the pvalue of Z when X = 50. So
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
By the Central Limit Theorem
![Z = \frac{X - \mu}{s}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7Bs%7D)
![Z = \frac{50 - 54}{1.8}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B50%20-%2054%7D%7B1.8%7D)
![Z = -2.22](https://tex.z-dn.net/?f=Z%20%3D%20-2.22)
has a pvalue of 0.0132.
1.32% probability that a randomly selected 100 customers will spend an average of less than $50 on dinner