Question

The number of diners at a restaurant each day is recorded and a daily average is calculated every month (assume 30 days in a month). The number of diners each day has a mean of 107 and a standard deviation of 60, but does not necessarily follow a normal distribution.The probability that a daily average over a given month is greater than x is 2.5%. Calculate x. You may find standard normal table useful. Give your answer to 3 decimal places.x =

Answer 1

Answer:

x = 128.472

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 normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

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 p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

The number of diners each day has a mean of 107 and a standard deviation of 60.

This means that $\mu = 107, \sigma = 60$

Distribution of the daily average:

Over a month of 30 days, so $n = 30, s = \frac{60}{\sqrt{30}} = 10.955$

The probability that a daily average over a given month is greater than x is 2.5%. Calculate x.

This is X when Z has a p-value of 1 - 0.025 = 0.975, so X when Z = 1.96. Then

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$1.96 = \frac{X - 107}{10.955}$

$X - 107 = 1.96*10.955$

$X = 128.472$

So x = 128.472