Question

Suppose that the IQs of university​ A's students can be described by a normal model with mean 150150 and standard deviation 77 points. Also suppose that IQs of students from university B can be described by a normal model with mean 120120 and standard deviation 1010. ​a) Select a student at random from university A. Find the probability that the​ student's IQ is at least 140140 points.

Answer 1

Answer:

The probability that the​ student's IQ is at least 140 points is of 55.17%.

Step-by-step 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:

$Z = \frac{X - \mu}{\sigma}$

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:

University A: $\mu = 150, \sigma = 77$

a) Select a student at random from university A. Find the probability that the​ student's IQ is at least 140 points.

This is 1 subtracted by the pvalue of Z when X = 140. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{140 - 150}{77}$

$Z = -0.13$

$Z = -0.13$ has a pvalue of 0.4483.

1 - 0.4483 = 0.5517

The probability that the​ student's IQ is at least 140 points is of 55.17%.