Question

a) Find P(X < 65)
b) P(X >= 72)
c) P(60 < X < 72)
d) How tall would you need to be to be among the tallest 5% of people in the school?

Answer 1

What's the mean/standard deviation? How is the population distributed?

Answer 2

@SithsAndGiggles
It is known that heights (X) at a local school follow a normal distribution with a mean of 68 inches and a standard deviation of 2.7 inches.

Answer 3

\[P(X<65)=P\left(\frac{X-68}{2.7}<\frac{65-68}{2.7}\right)=P(Z<-1.11)\]
\[P(X\ge72)=P\left(\frac{X-68}{2.7}\ge\frac{72-68}{2.7}\right)=P(Z\ge1.48)\]
\[P(60<X<72)=P\left(\frac{60-68}{2.7}<\frac{X-68}{2.7}<\frac{72-68}{2.7}\right)=P(-2.96<Z<1.48)\]
For (d), you want to find the height \(k\) such that
\[P(X>k)=0.05\]
which entails finding the proper z-score.
\[P\left(\frac{X-68}{2.7}>\frac{k-68}{2.7}\right)=0.05\]
For a probability of 0.05, the corresponding z-score is 1.645, which means you solve for \(k\) in
\[\frac{k-68}{2.7}=1.645\]

Answer 4

cutoff part: \(P(-2.96<Z<1.48)\)

Answer 5

@SithsAndGiggles Thank you very much, that makes much more sense now

Answer 6

@SithsAndGiggles wait but how do you know that the corresponding z-score is 1.645 again? Is that just from like a chart

Answer 7

Yes: http://www.stat.ufl.edu/~athienit/Tables/Ztable.pdf

In this table (a left-tail table), you have \(P(Z<-1.645)=0.05\), which - due to the symmetry of the curve - means \(P(Z>1.645)=0.05\).