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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?

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User Oob
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1 Answer

5 votes

Answer:

The bonus will be paid on at least 4099 units.

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


1.645 = (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.

answered
User James Marble
by
8.5k points

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