Question

The mean gas mileage for a hybrid car is 57 miles per gallon. Suppose that the gasoline mileage is approximately normally distributed with a standard deviation of 3.5 miles per gallon. ​

(a) What proportion of hybrids gets over 61 miles per​ gallon?

(b) What proportion of hybrids gets 52 miles per gallon or​ less?

(c) What proportion of hybrids gets between 59 and 62 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 = = 57

standard deviation = = 3.5

a ) P (x > 61)

= 1 - P (x < 61 )

= 1 - P ( x -  / ) < ( 61 - 57 / 3.5 )

= 1 - P ( z < 4 / 3.5 )

= 1 - P ( z < 1.14 )

Using z table

= 1 -0.8729

= 0.1271

Probability = 0.1271

b ) P( x < 52 )

P ( x - / ) < ( 52 - 57 / 3.5 )

P ( z < - 5 / 3.5 )

P ( z < -1.43)

= 0.0764

Probability =0.0764

c ) P (59 < x < 62 )

P ( 59 - 57 / 3.5) < ( x -  / ) < ( 62 -57 / 3.5)

P ( 2 / 3.5 < z < 5 / 3.5 )

P (0.57 < z < 1.43 )

P ( z < 1.43 ) - P ( z < 0.57 )

Using z table

= 0.9236 - 0.7157

= 0.2079

Probability = 0.2079

d ) P( x < 46 )

P ( x - / ) < ( 46 - 57 / 3.5 )

P ( z < -11 / 3.5 )

P ( z < -3.14 )

= 0.0008

Probability =0.0008