Answer:
57.62% probability that a randomly selected adult has an IQ between 89 and 121.
Explanation:
When the distribution is normal, we use the z-score formula.
In a set with mean
and standard deviation
, the zscore of a measure X is given by:
![Z = (X - \mu)/(\sigma)](https://img.qammunity.org/2021/formulas/mathematics/college/c62rrp8olhnzeelpux1qvr89ehugd6fm1f.png)
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 question, we have that:
![\mu = 105, \sigma = 20](https://img.qammunity.org/2021/formulas/mathematics/college/iibhyv6o7rlxv601rcsvij3ywt6fb0ezee.png)
Find the probability that a randomly selected adult has an IQ between 89 and 121.
This is the pvalue of Z when X = 121 subtracted by the pvalue of Z when X = 89. So
X = 121
![Z = (X - \mu)/(\sigma)](https://img.qammunity.org/2021/formulas/mathematics/college/c62rrp8olhnzeelpux1qvr89ehugd6fm1f.png)
![Z = (121 - 105)/(20)](https://img.qammunity.org/2021/formulas/mathematics/college/242vtnaxj0wp97grej55ew0bvvpiwsxlyi.png)
![Z = 0.8](https://img.qammunity.org/2021/formulas/mathematics/college/puk4mqa5csaqmsmyy66cywlbdurc2e9d2s.png)
has a pvalue of 0.7881
X = 89
![Z = (X - \mu)/(\sigma)](https://img.qammunity.org/2021/formulas/mathematics/college/c62rrp8olhnzeelpux1qvr89ehugd6fm1f.png)
![Z = (89 - 105)/(20)](https://img.qammunity.org/2021/formulas/mathematics/college/wpnlkj2n5bh9ypn4kaoickpiysvqg29ivh.png)
![Z = -0.8](https://img.qammunity.org/2021/formulas/mathematics/college/kkhnpkiefckabwmly12vsk03f6b1bmr4zt.png)
has a pvalue of 0.2119
0.7881 - 0.2119 = 0.5762
57.62% probability that a randomly selected adult has an IQ between 89 and 121.