Question

The spending at Target is distributed normally with a mean spending of $47.67 and a standard deviation of $5.50. What is the probability that the spending is between 46 and 49.56 dollars?

Answer 1

25.10% probability that the spending is between 46 and 49.56 dollars

Explanation:

Problems of normally distributed samples are solved using 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 problem, we have that:

`\mu = 47.67, \sigma = 5.5`

What is the probability that the spending is between 46 and 49.56 dollars?

This is the pvalue of Z when X = 49.56 subtracted by the pvalue of Z when X = 46. So

X = 49.56

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

`Z = (49.56 - 47.67)/(5.5)`

`Z = 0.34`

`Z = 0.34` has a pvalue of 0.6331

X = 46

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

`Z = (46 - 47.67)/(5.5)`

`Z = -0.3`

`Z = -0.3` has a pvalue of 0.3821

0.6331 - 0.3821 = 0.2510

25.10% probability that the spending is between 46 and 49.56 dollars