Question

The mean gas mileage for a hybrid car is 56 miles per gallon. Suppose that the gasoline mileage is approximately normally distributed with a standard deviation of 3.2 miles per gallon.​ (a) What proportion of hybrids gets over 61 miles per​ gallon? (b) What proportion of hybrids gets 51 miles per gallon or​ less? left parenthesis c right parenthesis What proportion of hybrids gets between 59 and 61 miles per​ gallon? (d) What is the probability that a randomly selected hybrid gets less than 46 miles per​ gallon?

Answer 1

Solution:

Given that,

mean = = 56

standard deviation = = 3.2

A ) p ( x > 61 )

= 1 - p (x < 61 )

= 1 - p ( x -  / ) < ( 61 - 56 / 3.2)

= 1 - p ( z < 5 / 3.2 )

= 1 - p ( z < 1.56)

Using z table

= 1 - 0.9406

= 0.0594

Probability = 0.0594

b ) p ( x < 51 )

= p ( x -  / ) < ( 51 - 56 / 3.2)

= p ( z < - 5 / 3.2 )

= p ( z < - 1.56)

Using z table

= 0.0594

Probability = 0.0594

C ) p ( 59 < x < 61)

= p ( 59 - 56 / 3.2) < ( x -  / ) < ( 61 - 56 / 3.2)

= p ( 3 / 3.2 < z < 5 / 3.2 )

= p ( 0.94 < z < 1.56 )

= p (z < 1.56 ) - p ( z < 0.94 )

Using z table

= 0.9406 - 0.8264

= 0.1142

Probability = 0.1142

d ) p ( x < 46 )

= p ( x -  / ) < ( 46 - 56 / 3.2)

= p ( z < - 10 / 3.2 )

= p ( z < - 3.13)

Using z table

= 0.0009

Probability = 0.0009