Question

Suppose that the IQ of a randomly selected student from a university is normal with mean 115 and standard deviation 25. Determine the interval of values that is centered at the mean and for which 50% of the students have IQ's in that interval.

Answer 1

The interval is [98,132]

Explanation:

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Normal with mean 115 and standard deviation 25.

This means that
`\mu = 115, \sigma = 25`

Determine the interval of values that is centered at the mean and for which 50% of the students have IQ's in that interval.

Between the 50 - (50/2) = 25th percentile and the 50 + (50/2) = 75th percentile.

25th percentile:

X when Z has a p-value of 0.25, so X when Z = -0.675.

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

`-0.675 = (X - 115)/(25)`

`X - 115 = -0.675*25`

`X = 98`

75th percentile:

X when Z has a p-value of 0.75, so X when Z = 0.675.

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

`0.675 = (X - 115)/(25)`

`X - 115 = 0.675*25`

`X = 132`

The interval is [98,132]