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
and standard deviation
, the zscore of a measure X is given by
![Z = (X - \mu)/(\sigma)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/ijf8wrxup4oiph7gw8zex0r9316mpsigqy.png)
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
.
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](https://img.qammunity.org/2020/formulas/mathematics/college/464rdl7nm8fpk2yew7mlhf9d617wxr9f15.png)
![Z = (X - \mu)/(\sigma)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/ijf8wrxup4oiph7gw8zex0r9316mpsigqy.png)
![1.645 = (X - 4000)/(60)](https://img.qammunity.org/2020/formulas/mathematics/college/pmez0c205apd60go88yakd4y9u5i6n1u5r.png)
![X - 4000 = 60*1.645](https://img.qammunity.org/2020/formulas/mathematics/college/ayku442g9qm2nwo4tey18jqyhoacclcvxg.png)
![X = 4098.7](https://img.qammunity.org/2020/formulas/mathematics/college/5gj6w9pxae4m0zv0ne6lzfzjobs0mpnvju.png)
The number of units is discrete, this means that the bonus will be paid on at least 4099 units.