Question

Companies who design furniture for elementary school classrooms produce a variety of sizes for kids of different ages. Suppose the heights of kindergarten children can be described by a Normal model with a mean of 39.2

inches and standard deviation of

1.9inches.

​a) What fraction of kindergarten kids should the company expect to be less than

33 inches​ tall?About blank ​% of kindergarten kids are expected to be less than 33 inches tall.

​(Round to one decimal place as​ needed.)

​b) In what height interval should the company expect to find the middle 80​% of​ kindergarteners?The middle 80​% of kindergarteners are expected to be between what inches and what inches.

​(Use ascending order. Round to one decimal place as​ needed.)

​c) At least how tall are the biggest 30​% of​ kindergarteners?The biggest 30​% of kindergarteners are expected to be at least ? inches tall.

​(Round to one decimal place as​ needed.)

Answer 1

X : Height of kindergarten children

X follows normal distribution with mean 39.2 inches and standard deviation of 1.9 inches.

a) Probability that a kindergarten kid's height less than 33 inches​ =P(X<33)

Z-score for 33 = (33-39.2)/1.9 = -3.26

From standard normal tables P(Z<-3.26) = 0.0006

P(X<33) = P(Z<-3.26) = 0.0006

Probability that a kindergarten kid's height less than 33 inches​ =P(X<33) =0.0006

% of kindergarten kids are expected to be less than 33 inches tall = 0.0006x100=0.06%

About 0.1%(or 0.06%) of kindergarten kids are expected to be less than 33 inches tall.

b) In what height interval should the company expect to find the middle 80​% of​ kindergarteners

Middle : 80% means ; Left tail : 10% and right tail :10%

i.e Let the range be X_L : Lower ; X_H : Higher value

P(X_L < X < X_H) =0.80

Left Tail : P(X<X_L) = 0.10; Right Tail :P(X>X_H) = 0.10 ; P(X>X_H) =1-P(X<X_H) = 0.10 ; P(X<X_H) = 1-0.10=0.90

Z_L = Z-score for X_L = (X_L - 39.2)/1.9 ; Z_H = Z-score for X_H = (XH - 39.2)/1.9

X_L = 39.2 + 1.9Z_L ; X_H = 39.2 + 1.9Z_H

P(Z<Z_L) =  P(X<X_L) = 0.10 ; P(Z<Z_H) =  P(X<X_H) = 0.90

From standard normal tables.

P(Z<-1.28) =​​​ 0.1003; P(Z<1.28) = 0.8997

Z_L = -1.28 ; Z_H = 1.28

X_L = 39.2 + 1.9Z_L = 39.2 + 1.9 x (-1.28)=36.768 ; X_H = 39.2 + 1.9Z_H = 39.2 + 1.9 x (+1.28)=41.632

The middle 80​% of kindergartners are expected to be between 36.768 inches and 41.632 inches

​c) At least how tall are the biggest 30​% of ​kindergarteners

Let X₃ such that P(X>X₃) =30/100 =0.3

P(X>X₃) =0.3

P(X>X₃) = 1- P(X<X₃) = 0.3

P(X<X₃) = 1-0.3 =0.7

Z₃ : Z score for X₃ = (X₃ - 39.2)/1.9

X₃ = 39.2 + 1.9Z₃

P(X<X₃) = P(Z<Z₃) =0.7

From standard normal tables,

P(Z<0.53) =0.7019

Z₃ =0.53

X₃ = 39.2 + 1.9Z₃ = 39.2 + 1.9 x 0.53 =40.207

The biggest 30​% of kindergarteners are expected to be at least 40.2 inches tall