Question

According to the “Bottled Water Trends for 2014” report (bit.ly/1gx5ub8), the U.S. per capita consumption of bottled water in 2013 was 31.8 gallons. Assume that the per capita consumption of bottled water in the United States is approximately normally distributed with a mean of 31.8 gallons and a standard deviation of 10 gallons.

PLEASE USE NORMDIST AND NORMINV IN EXCEL

a. What is the probability that someone in the United States con- sumed more than 32 gallons of bottled water in 2013?

b. What is the probability that someone in the United States consumed between 10 and 20 gallons of bottled water in 2013?

c. What is the probability that someone in the United States consumed less than 10 gallons of bottled water in 2013?

d. Ninety-nine percent of the people in the United States consumed less than how many gallons of bottled water?

e) The amount of water consumed by 92% of US population will be between what two values symmetrically distributed around the mean?

Answer 1

Part a)

X ~ N ( µ = 31.8 , σ = 10 )
P ( X > 32 ) = 1 - P ( X < 32 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 32 - 31.8 ) / 10
Z = 0.02
P ( ( X - µ ) / σ ) > ( 32 - 31.8 ) / 10 )
P ( Z > 0.02 )
P ( X > 32 ) = 1 - P ( Z < 0.02 )
P ( X > 32 ) = 1 - 0.508
P ( X > 32 ) = 0.492

1-NORMSDIST( Z = 0.02 )

Part b)

X ~ N ( µ = 31.8 , σ = 10 )
P ( 10 < X < 20 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 10 - 31.8 ) / 10
Z = -2.18
Z = ( 20 - 31.8 ) / 10
Z = -1.18
P ( -2.18 < Z < -1.18 )
P ( 10 < X < 20 ) = P ( Z < -1.18 ) - P ( Z < -2.18 )
P ( 10 < X < 20 ) = 0.119 - 0.0146
P ( 10 < X < 20 ) = 0.1044

NORMSDIST( Z = -1.18) - NORMSDIST(Z = -2.18 )

Part c)

X ~ N ( µ = 31.8 , σ = 10 )
P ( X < 10 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 10 - 31.8 ) / 10
Z = -2.18
P ( ( X - µ ) / σ ) < ( 10 - 31.8 ) / 10 )
P ( X < 10 ) = P ( Z < -2.18 )
P ( X < 10 ) = 0.0146

NORMSDIST( Z = -2.18 )

Part d)

X ~ N ( µ = 31.8 , σ = 10 )
P ( X < ? ) = 99% = 0.99
Looking for the probability 0.99 in standard normal table to calculate critical value Z = 2.3263
Z = ( X - µ ) / σ
2.3263 = ( X - 31.8 ) / 10
X = 55.063
P ( X < 55.063 ) = 0.99

NORMSINV( 0.99 )

part e)

X ~ N ( µ = 31.8 , σ = 10 )
P ( a < X < b ) = 0.92
Dividing the area 0.92 in two parts we get 0.92/2 = 0.46
since 0.5 area in normal curve is above and below the mean
Area below the mean is a = 0.5 - 0.46
Area above the mean is b = 0.5 + 0.46
Looking for the probability 0.04 in standard normal table to calculate critical value Z = -1.7507
Looking for the probability 0.96 in standard normal table to calculate critical value Z = 1.7507
Z = ( X - µ ) / σ
-1.7507 = ( X - 31.8 ) / 10
a = 14.293
1.7507 = ( X - 31.8 ) / 10
b = 49.307
P ( 14.293 < X < 49.307 ) = 0.92

NORMSINV( 0.04 )

NORMSINV( 0.96 )