Question #2373

In a study for Health Statistics, it was found that the mean height of female 20-25 years old was 64.1 inches. If height is normally distributed with a standard deviation of 2.8 inches.
a) Determine the height required to be in the top 10% of all 20-25 years old females.
b) A “one-size-fits all” robe is being designed that should fit 90% of 20-25 years old females; what heights constitute the middle 90% of all 20-25 years old females?

Expert's answer

Let X be height of female 20-25 years old, then Z = (X-64.1)/2.8 has standard normal distribution.

a) We have to such H that

P(X>H)=0.1

0.1 = P(X>H) = P( Z>(H-64.1)/2.8 ) =

= 1 - F((H-64.1)/2.8)

So

F((H-64.1)/2.8) = 0.9

Then, using known values of F we obtain that

F((H-64.1)/2.8)= 1.281552

Hence

H =1.281552 * 2.8 + 64.1 = 67.69

b) We have to find D such that

P(64.1-D < X < 64.1+D ) = 0.9

P(64.1-D < X < 64.1+D )& =

= P(-D/2.8 < Z < D/2.8 ) =

= F(D/2.8) - F(-D/2.8) =

= F(D/2.8) - (1-F(D/2.8))& =

= 2 * F(D/2.8) – 1 = 0.9

& F(D/2.8) = 1.9/2 = 0.95

& D/2.8 = 1.644854

D = 4.6

Hence the heights being in the middle 90% of all 20-25 years old females should be in the following interval:

& [ 64.1-4.6, 64.1+4.6 ]& =& [59.5,& 68.7]

a) We have to such H that

P(X>H)=0.1

0.1 = P(X>H) = P( Z>(H-64.1)/2.8 ) =

= 1 - F((H-64.1)/2.8)

So

F((H-64.1)/2.8) = 0.9

Then, using known values of F we obtain that

F((H-64.1)/2.8)= 1.281552

Hence

H =1.281552 * 2.8 + 64.1 = 67.69

b) We have to find D such that

P(64.1-D < X < 64.1+D ) = 0.9

P(64.1-D < X < 64.1+D )& =

= P(-D/2.8 < Z < D/2.8 ) =

= F(D/2.8) - F(-D/2.8) =

= F(D/2.8) - (1-F(D/2.8))& =

= 2 * F(D/2.8) – 1 = 0.9

& F(D/2.8) = 1.9/2 = 0.95

& D/2.8 = 1.644854

D = 4.6

Hence the heights being in the middle 90% of all 20-25 years old females should be in the following interval:

& [ 64.1-4.6, 64.1+4.6 ]& =& [59.5,& 68.7]

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