Scores of IQ test have bell-shaped distribution with a mean of 100 and a standard deviation of 18. Use the empirical rule to determine the following. A. What percentage of people has an IQ score between 46 and 154?
B. What percentage of people has an IQ scores less than 82 or greater than 118? C. What percentage is greater than 118?

Since [tex](46,154)=(100-3\times18,100+3\times18)[/tex], you have

[tex]\mathbb P(46<X<154)=\mathbb P(-3<Z<3)\approx99.7\%[/tex]

Since [tex](-\infty,82)\cup(118,\infty)=(-\infty,100-1\times18)\cup(100+1\times18,\infty)[/tex], you have

[tex]\mathbb P((X<82)\cup(X>118))=\mathbb P(Z<1)+\mathbb P(Z>1)=1-\mathbb P(-1<Z<1)[/tex]

The empirical rule says that approximately 68% of a normal distribution falls within one standard deviation of the mean, you have

[tex]100\%-\mathbb P(-1<Z<1)\approx68\%\implies \mathbb P(-1<Z<1)\approx32\%[/tex]

Finally, since those with IQ greater than 118 comprise half of the above percentage (due to the distribution's symmetry), you have

[tex]\mathbb P(Z>1)\approx50\%\times32\%=16\%[/tex]

kobenhavn kobenhavn · Answer 2 · 2021-09-29T13:44:04+02:00

[tex]99.7[/tex]% of people has an IQ score between [tex]46[/tex] and [tex]154[/tex].

[tex]32[/tex]% of people has an IQ scores less than [tex]82[/tex] or greater than [tex]118[/tex].

[tex]16[/tex]% percentage is greater than [tex]118[/tex].

A) Mean [tex]= 100[/tex]

Standard deviation [tex]= 18[/tex]

IQ between [tex]46[/tex] and [tex]154[/tex]

For [tex](46,154) = (100-3\times 18, 100+3\times 18)[/tex]

[tex]P(46<X<154)[/tex]

[tex]=P(-3<Z<3) \approx 99.7%[/tex]%

B)

[tex](-\infty, 82)U(118, \infty)=(-\infty, 100-1\times 18)U(100+1\times 18, \infty)[/tex]

[tex]P((X<82)U(X>118))=P(Z<1)+P(Z>1)[/tex]

[tex]=1-P(-1<Z<1)[/tex]

According to the empirical rule, approximately [tex]68[/tex]% of a normal distribution falls within one standard deviation of the mean

[tex]100[/tex]%[tex]-P(-1<Z<1)\approx 68[/tex]%

[tex]P(-1<Z<1)\approx 32[/tex]%

C)

Now, since those with IQ greater than [tex]118[/tex] comprise half of the above percentage (due to the distribution's symmetry),

[tex]P(Z>1)\approx 50[/tex]% [tex]\times 32[/tex]% [tex]=16[/tex]%

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