Question #290410

Given a standard normal distribution with mean =78 and standard deviation = 28, find P(85 < X < 102)


1
Expert's answer
2022-01-25T14:47:21-0500

Given: μ=78,σ=28\mu=78,\sigma=28

P(85<X<102)=P(X<102)P(X<85)P(85 < X < 102)=P(X<102)-P(X<85)

Now,

P[x<102]=P[z<xμσ]=P[z<1027828]=P[z<0.857]=0.80428\begin{aligned} & P[x <102] \\ =& P\left[z <\frac{x-\mu}{\sigma}\right] \\ =& P\left[z <\frac{102-78}{28}\right] \\ =& P[z <0.857] \\ =& 0.80428 \\ \end{aligned}

and

P[x<85]=P[z<xμσ]=P[z<857828]=P[z<0.25]=0.59871\begin{aligned} & P[x <85] \\ =& P\left[z <\frac{x-\mu}{\sigma}\right] \\ =& P\left[z <\frac{85-78}{28}\right] \\ =& P[z <0.25] \\ =& 0.59871 \\ \end{aligned}

So,

P(85<X<102)=P(X<102)P(X<85)=0.804280.59871=0.20557P(85 < X < 102)=P(X<102)-P(X<85)=0.80428-0.59871=0.20557


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