Question

The distribution of students’ heights in a class of 100 students is normal, with a mean height of 66 inches and a standard deviation of three. With these parameters, answer the associated question(s). Between which two heights (in inches) do the middle 60 students fall? Round to the nearest tenths place if a fraction.

Answer 1

The middle 60 students fall between 63.48 inches and 68.52 inches.

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.

In this problem, we have that:

`\mu = 66, \sigma = 3`

Between which two heights (in inches) do the middle 60 students fall?

The normal probability distribution is symmetric. So the middle 60% fall from a pvalue of 0.50 - 0.60/2 = 0.20(lower bound) to a pvalue of 0.50 + 0.60/2 = 0.80(upper bound)

Lower bound

X when Z has a pvalue of 0.20.

So X when
`Z = -0.84`

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

`-0.84 = (X - 66)/(3)`

`X - 66 = -0.84*3`

`X = 63.48`

Upper bound

X when Z has a pvalue of 0.80.

So X when
`Z = 0.84`

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

`0.84 = (X - 66)/(3)`

`X - 66 = 0.84*3`

`X = 68.52`

The middle 60 students fall between 63.48 inches and 68.52 inches.