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?
Answers
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.