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: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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by[tex]Z = \frac{X - \mu}{\sigma}[/tex]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 [tex]\mu = 4000, \sigma = 60[/tex].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 [tex]Z = 1.645[/tex][tex]Z = \frac{X - \mu}{\sigma}[/tex][tex]1.645 = \frac{X - 4000}{60}[/tex][tex]X - 4000 = 60*1.645[/tex][tex]X = 4098.7[/tex]The number of units is discrete, this means that the bonus will be paid on at least 4099 units.