Given a standard normal distribution with mean =78 and standard deviation = 28, find P(85 < X < 102)
Given: μ=78,σ=28\mu=78,\sigma=28μ=78,σ=28
P(85<X<102)=P(X<102)−P(X<85)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<x−μσ]=P[z<102−7828]=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}====P[x<102]P[z<σx−μ]P[z<28102−78]P[z<0.857]0.80428
and
P[x<85]=P[z<x−μσ]=P[z<85−7828]=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}====P[x<85]P[z<σx−μ]P[z<2885−78]P[z<0.25]0.59871
So,
P(85<X<102)=P(X<102)−P(X<85)=0.80428−0.59871=0.20557P(85 < X < 102)=P(X<102)-P(X<85)=0.80428-0.59871=0.20557P(85<X<102)=P(X<102)−P(X<85)=0.80428−0.59871=0.20557
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