Question

Management is considering adopting a bonus system to increase production. One suggestion is to pay a bonus on the highest 5 percent of production based on past experience. Past records indicate that, on the average, 4,000 units of a small assembly are produced during a week. The distribution of the weekly production is approximately normally distributed with a standard deviation of 60 units. If the bonus is paid on the upper 5 percent of production, the bonus will be paid on how many units or more?

Answer 1

Answer:

The bonus will be paid on at least 4099 units.

Step-by-step explanation:

Problems of normally distributed samples 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

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

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 percentile of this measure.

In this problem, we have that:

The highest 5 percent is the 95th percentile.

Past records indicate that, on the average, 4,000 units of a small assembly are produced during a week. The distribution of the weekly production is approximately normally distributed with a standard deviation of 60 units. This means that $\mu = 4000, \sigma = 60$.

If the bonus is paid on the upper 5 percent of production, the bonus will be paid on how many units or more?

The least units that the bonus will be paid is X when Z has a pvalue of 0.95.

Z has a pvalue of 0.95 between 1.64 and 1.65. So we use $Z = 1.645$

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

$1.645 = \frac{X - 4000}{60}$

$X - 4000 = 60*1.645$

$X = 4098.7$

The number of units is discrete, this means that the bonus will be paid on at least 4099 units.