Question

A market research company employs a large number of typists to enter data into a computer database. The time it takes for potential 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. A candidate is automatically hired if she learns the computer system in less than 100 minutes. A cut-off time is set at the slowest 10% of the learning distribution. Anyone slower than this cut-off time is definitely not hired. What proportion of candidates takes more than two hours to learn the computer system?

Answer 1

Answer:

• 4.75% of the candidates takes more than two hours to learn the computer system.

Explanation:

The relevant information to solve the problem is:

• 1. The time it takes to learn follows a Normal distribution

• 2. The mean is 90 minutes

• 3. The standard deviation is 18 minutes

• 4. The question is What proportion of candidates takes more than two hours to learn the computer system?

Then, you shall calculate the Z-score and use a standard distribution table to look up the Z-score and the corresponding probability.

Repeating myself from a recent answer, "there are two types of standard distribution tables: tables that show values that represent the AREA to the LEFT of the Z-score, and tables that show values that represent the AREA to the RIGHT of the Z-score".

1. First, calculate the Z-score:

       $Z-score=\frac{x-mean}{standard\text{ }deviation}$

                          $x=2hours=120min$

        $Z-score=\frac{120-90}{18}\approx 1.67$

2. Use the table that represents the area to the right of the mean to find the ratio of typists that have a Z-score greater than 1.67.

          $Probability=0.0475=4.75\%$

Therefore, 4.75% of the candidates takes more than two hours to learn the computer system.