Question

Management of a company 1s considering adopting a
bonus system to increases production. One suggestion
1s to pay a bonus on the highest 5 per cent ot production
based on past experience. Past records indicate that,
on the average, 4000 units of a small assembly are
produced during a week. The distribution of the
weekly production is approximately normal with a
standard deviation of 60 units. If the bonus is paid on
the upper 5 per cent 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 \muμ and standard deviation \sigmaσ , the zscore of a measure X is given by

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

σ

X−μ

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 = 60μ=4000,σ=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 = 1.645Z=1.645

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

σ

X−μ

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

60

X−4000

X - 4000 = 60*1.645X−4000=60∗1.645

X = 4098.7X=4098.7

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