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.

The interval is [98,132]

Step-by-step explanation:

Normal Probability Distribution

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

In a set with mean and standard deviation , the z-score of a measure X is given by:

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

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.

75th percentile:

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

The interval is [98,132]

answer: idk

step-by-step explanation:

step-by-step explanation:

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