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Prusswan · Answer 1 · 2023-02-17T23:23:31+0000

Solution

We are given

$\begin{gathered} Mean(\mu)=100 \\ S\tan dardDeviation(\sigma)=10 \end{gathered}$

Note: Z score formula

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

(a). What percentage of people has an IQ score between 90 and 110?

$\begin{gathered} p(90Therefore, the answer is[tex]p(90(b) What percentage of people has an IQ score less than 80 or greater than 120?Thus, we have[tex]\begin{gathered} p(X<80)+p(X>120) \\ \text{For } \\ p(X<80)=p(Z<(80-100)/(10)) \\ p(X<80)=p(Z<-2) \\ p(X<80)=0.02275 \\ \text{Now for} \\ p(X>120)=p(Z>(120-100)/(10)) \\ p(X>120)=p(Z>2)=1-p(Z<2)=1-0.97725=0.02275 \\ \text{Therefore,} \\ p(X<80)+p(X>120)=0.02275+0.02275 \\ p(X<80)+p(X>120)=0.0455 \\ \text{Converting to percentage} \\ p(X<80)+p(X>120)=4.55\% \end{gathered}$

Therefore, the answer is

$p(X<80)+p(X>120)=4.55\%$

(C). What percentage of people has an IQ score greater than 120?

We have

$\begin{gathered} p(X>120)=p(Z>(120-100)/(10)) \\ p(X>120)=p(Z>2)=1-p(Z<2)=1-0.97725=0.02275 \\ \text{Converting to percentage} \\ p(X>120)=2.275\% \end{gathered}$

Therefore, the answer is

$p(X>120)=2.275\%$

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