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Sagot :
To solve this question, we will have to find the Z score.
[tex]Z=\frac{x-\bar{x}}{s.d}[/tex]Where x is the value of the IQ
X-bar is the mean.
s.d is the standard deviation
[tex]\begin{gathered} Z_1=\frac{82-100}{18} \\ =-\frac{18}{18} \\ =-1 \end{gathered}[/tex][tex]Z_2=\frac{100-100}{18}[/tex][tex]\begin{gathered} Z_2=\frac{0}{18} \\ =0 \end{gathered}[/tex]The percentage of the population with IQs between 82 and 100 will be calculated thus:
[tex]\begin{gathered} P(-1-1) \\ =0.5-0.1587 \\ =0.34134 \\ \text{The percentage is}\colon \\ =0.34134\times100=34.134\text{ \%} \end{gathered}[/tex]
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