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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A. B. C. D. E.D
We want the area 1.0 - 0.55 = 0.45. For this we have a z value of 1.645. Now, using z = (x-u)/sigma, we have 1.645 = (x-4000)/60. x = 4099.
To determine the bonus threshold for the highest 5 percent of production, we need to find the production level that corresponds to the 95th percentile of the production distribution.
Given that the average production is 4,000 units and the standard deviation is 60 units, we can use these parameters to calculate the z-score corresponding to the 95th percentile.
The z-score formula is:
z = (x - μ) / σ
where:
To find the z-score at the 95th percentile, we need to find the z-score such that the area under the normal distribution curve to the left of that z-score is 0.95.
Using a standard normal distribution table or a statistical calculator, we can find that the z-score for a cumulative probability of 0.95 is approximately 1.645.
Now, we can solve for x in the z-score formula:
1.645 = (x - 4,000) / 60
Simplifying the equation:
1.645 * 60 = x - 4,000
98.7 = x - 4,000
x = 98.7 + 4,000
x ≈ 4,098.7
Rounding to the nearest whole number, the bonus will be paid on 4,099 units or more.
Therefore, the correct answer is D. 4099.