Question

The manager of a new supermarket wished to estimate the likely expenditure of his customers. A sample of till slips from a similar supermarket describing the weekly amount spent by 500 randomly selected customers was collected and analysed. This expenditure was found to be approximately normally distributed with a mean of $50 and a standard deviation of $15.
Find the probability that any shopper selected at random spends more than $80 per week?
Find the percentage of shoppers who are expected to spend between $30 and 80 per week?

Answer 1

Answer:

0.0228 = 2.28% probability that any shopper selected at random spends more than $80 per week.

88.54% of shoppers are expected to spend between $30 and 80 per week.

Step-by-step explanation:

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.

Normally distributed with a mean of $50 and a standard deviation of $15.

This means that $\mu = 50, \sigma = 15$

Find the probability that any shopper selected at random spends more than $80 per week?

This is 1 subtracted by the p-value of Z when X = 80. So

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

$Z = \frac{80 - 50}{15}$

$Z = 2$

$Z = 2$ has a p-value of 0.9772

1 - 0.9772 = 0.0228

0.0228 = 2.28% probability that any shopper selected at random spends more than $80 per week.

Find the percentage of shoppers who are expected to spend between $30 and 80 per week?

The proportion is the p-value of Z when X = 80 subtracted by the p-value of Z when X = 30.

X = 80

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

$Z = \frac{80 - 50}{15}$

$Z = 2$

$Z = 2$ has a p-value of 0.9772

X = 30

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

$Z = \frac{30 - 50}{15}$

$Z = -1.33$

$Z = -1.33$ has a p-value of 0.0918

0.9772 - 0.0918 = 0.8854

0.8854*100% = 88.54%

88.54% of shoppers are expected to spend between $30 and 80 per week.