Question

 

A recent survey shows that the average of American household’s annual expenditure on coffee is $54.16 with a standard deviation of $10.50. We randomly select a household. Assuming that the expenditure follows normal distribution, find:

a)The probability that their annual expenditure on coffee is less than $25?

b)The probability that it is more than $50?

c)The probability that it is between $40 and $60?

d)The amount less than which 95% of households spend on coffee?

Answer 1

NORM.DIST(x,mean,standard_dev,cumulative)

Cumulative :- When we need exact probability, cumulative would be false, else it will always be true.

Part a)
X ~ N ( µ = 54.16 , σ = 10.5 )
P ( X < 25 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 25 - 54.16 ) / 10.5
Z = -2.7771
P ( ( X - µ ) / σ ) < ( 25 - 54.16 ) / 10.5 )
P ( X < 25 ) = P ( Z < -2.7771 )
P ( X < 25 ) = 0.0027

Excel function - NORM.DIST(25,54.16,10.5,TRUE) = 0.0027419

Part b)
X ~ N ( µ = 54.16 , σ = 10.5 )
P ( X > 50 ) = 1 - P ( X < 50 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 50 - 54.16 ) / 10.5
Z = -0.3962
P ( ( X - µ ) / σ ) > ( 50 - 54.16 ) / 10.5 )
P ( Z > -0.3962 )
P ( X > 50 ) = 1 - P ( Z < -0.3962 )
P ( X > 50 ) = 1 - 0.346
P ( X > 50 ) = 0.654

Excel function - 1 - NORM.DIST(50,54.16,10.5,TRUE) = 0.654017

Part c)
X ~ N ( µ = 54.16 , σ = 10.5 )
P ( 40 < X < 60 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 40 - 54.16 ) / 10.5
Z = -1.3486
Z = ( 60 - 54.16 ) / 10.5
Z = 0.5562
P ( -1.35 < Z < 0.56 )
P ( 40 < X < 60 ) = P ( Z < 0.56 ) - P ( Z < -1.35 )
P ( 40 < X < 60 ) = 0.711 - 0.0887
P ( 40 < X < 60 ) = 0.6222

Excel function
NORM.DIST(40,54.16,10.5,TRUE) = 0.088737
NORM.DIST(60,54.16,10.5,TRUE) = 0.710995

Required probability = 0.710995 - 0.088737 = 0.622222

Part d)

X ~ N ( µ = 54.16 , σ = 10.5 )
P ( X < x ) = 95% = 0.95
To find the value of x
Looking for the probability 0.95 in standard normal table to calculate critical value Z = 1.6449
Z = ( X - µ ) / σ
1.6449 = ( X - 54.16 ) / 10.5
X = 71.4314
P ( X < 71.4314 ) = 0.95

Excel function -  NORM.INV(0.95,54.16,10.5) = 71.43096