Question

The daily dinner bills in a local restaurant are normally distributed with a mean of $28 and a standard deviation of $6. What are the minimum value of the bill that is greater than 95% of the bills?

Answer 1

The minimum value of the bill that is greater than 95% of the bills is $37.87.

Explanation:

When the distribution is normal, we use the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the zscore of a measure X is given by:

`Z = (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 pvalue, we get the probability that the value of the measure is greater than X.

In this question, we have that:

`\mu = 28, \sigma = 6`

What are the minimum value of the bill that is greater than 95% of the bills?

This is the 95th percentile, which is X when Z has a pvalue of 0.95. So X when Z = 1.645.

`Z = (X - \mu)/(\sigma)`

`1.645 = (X - 28)/(6)`

`X - 28 = 6*1.645`

`X = 37.87`

The minimum value of the bill that is greater than 95% of the bills is $37.87.