Question

Suppose that 20% of all invoices are for amounts greater than $1, 000. A random sample of 60 invoices is taken. What is the probability that the proportion of invoices in the sample is less than 18%?

Answer 1

34.83% probability that the proportion of invoices in the sample is less than 18%

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.

For a proportion p in a sample of size n, we have that
`\mu = p, \sigma = \sqrt{(p(1-p))/(n)}`

In this problem, we have that:

`p = 0.2, n = 60`

So

`\mu = 0.2, \sigma = \sqrt{(0.2*0.8)/(60)} = 0.0516`

What is the probability that the proportion of invoices in the sample is less than 18%?

This is the pvalue of Z when X = 0.18. So

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

`Z = (0.18 - 0.2)/(0.0516)`

`Z = -0.39`

`Z = -0.39` has a pvalue of 0.3483

34.83% probability that the proportion of invoices in the sample is less than 18%