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

39.239.2

inches and standard deviation of

1.9

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

3737

inches​ tall?About

of kindergarten kids are expected to be less than

inches tall.

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

​b) In what height interval should the company expect to find the middle

70​%

of​ kindergarteners?The middle

70​%

of kindergarteners are expected to be between

inches and

inches.

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

​c) At least how tall are the biggest

20​%

of​ kindergarteners?The biggest 20​%

of kindergarteners are expected to be at least

inches tall.

Answer 1

Part a)

X ~ N ( µ = 39.2 , σ = 1.9 )
P ( X < 37 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 37 - 39.2 ) / 1.9
Z = -1.16
P ( ( X - µ ) / σ ) < ( 37 - 39.2 ) / 1.9 )
P ( X < 37 ) = P ( Z < -1.16 )
P ( X < 37 ) = 0.123   0.1

Part b)

X ~ N ( µ = 39.2 , σ = 1.9 )
P ( a < X < b ) = 0.7
Dividing the area 0.7 in two parts we get 0.7/2 = 0.35
since 0.5 area in normal curve is above and below the mean
Area below the mean is a = 0.5 - 0.35
Area above the mean is b = 0.5 + 0.35
Looking for the probability 0.15 in standard normal table to calculate critical value Z = -1.04
Looking for the probability 0.85 in standard normal table to calculate critical value Z = 1.04
Z = ( X - µ ) / σ
-1.04 = ( X - 39.2 ) / 1.9
a = 37.224
1.04 = ( X - 39.2 ) / 1.9
b = 41.176
P ( 37.2 < X < 41.2 ) = 0.7

Part c)

X ~ N ( µ = 39.2 , σ = 1.9 )
P ( X > ? ) = 1 - P ( X < ? ) = 1 - 0.2 = 0.8
Looking for the probability 0.8 in standard normal table to calculate critical value Z = 0.84
Z = ( X - µ ) / σ
0.84 = ( X - 39.2 ) / 1.9
X = 40.8
P ( X > 40.8 ) = 0.2