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3. The heights of all adults in a large city have an approximately normal distribution with a mean of 68 inches and a standar
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Answer #1

Let , X\to N(\mu=68,\sigma=4)

a) Now ,

P(X<66)=P(\frac{X-\mu}{\sigma}<\frac{66-68}{4})=P(Z<-0.5)=\Phi(-0.5)

=1-\Phi(0.5)=1-0.6915=0.3085 ; From standard normal distribution table

b) Now , the sampling distribution of the sample mean is ,

\bar{X}\to N(\mu_{\bar{X}}=\mu,\sigma_{\bar{X}}=\frac{\sigma}{\sqrt{n}})

Therefore ,

P(67.5<\bar{X}<69)=P(\frac{67.5-68}{4/\sqrt{100}}<\frac{\bar{X}-\mu_{\bar{X}}}{\sigma_{\bar{X}}}<\frac{69-68}{4/\sqrt{100}})

=P(-1.25<Z<2.50)=P(Z<2.50)-P(Z<-1.25)

=\Phi(2.50)-\Phi(-1.25)=\Phi(2.50)-[1-\Phi(1.25)]

=0.9938-1+0.8944=0.8882 ; From standard normal distribution table

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