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

Answer:

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

Step-by-step 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:

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

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 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.

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

$1.645 = \frac{X - 28}{6}$

$X - 28 = 6*1.645$

$X = 37.87$

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