Question

The weights of certain machine components are normally distributed with a mean of 4.81 ounces and a standard deviation of 0.04 ounces. Find the two weights that separate the top 6% and the bottom 6%. These weights could serve as limits used to identify which components should be rejected. Round your answer to the nearest hundredth, if necessary.

Suppose SAT Writing scores are normally distributed with a mean of 496 and a standard deviation of 109. A university plans to award scholarships to students whose scores are in the top 7%. What is the minimum score required for the scholarship? Round your answer to the nearest whole number, if necessary.

Answer 1

First question:

Top 6%: 4.87 ounces

Bottom 6%: 4.75 ounces

Second question:

Top 7%: Score of 649.4.

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)`

The Z-score measures how many standard deviations the measure is from the mean. 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 probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

For the first problem, we have that:

`\mu = 4.81, \sigma = 0.04`

Top 6%

The value of X when Z has a pvalue of 0.94. This is
`Z = 1.555`

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

`1.555 = (X - 4.81)/(0.04)`

`X - 4.81 = 1.555*0.04`

`X = 4.8722`

Bottom 6%

The value of X when Z has a pvalue of 0.06. This is
`Z = 1.555`

For the second problem, we have that:

`\mu = 496, \sigma = 109`

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

`-1.555 = (X - 4.81)/(0.04)`

`X - 4.81 = -1.555*0.04`

`X = 4.7477`

Top 7%

The value of X when Z has a pvalue of 0.93. This is
`Z = 1.475`

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

`1.475 = (X - 496)/(104)`

`X - 496 = 104*1.475`

`X = 649.4`