Question

A market research company employs a large number of typists to enter data into a computer. The time taken for new typists to learn the computer system is known to have a normal distribution with a mean of 90 minutes and a standard deviation of 18 minutes. The proportion of new typists that take at most two hours to learn the computer system is

Answer 1

The proportion of new typists that take at most two hours to learn the computer system is 0.9525 = 95.25%.

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 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.

Normal distribution with a mean of 90 minutes and a standard deviation of 18 minutes.

This means that
`\mu = 90, \sigma = 18`

The proportion of new typists that take at most two hours to learn the computer system is

This is the pvalue of Z when X = 120 minutes = 2 hours. So

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

`Z = (120 - 90)/(18)`

`Z = 1.67`

`Z = 1.67` has a pvalue of 0.9525.

The proportion of new typists that take at most two hours to learn the computer system is 0.9525 = 95.25%.