Question

# The EPA Fuel Economy Estimates for automobile models tested recently predicted a mean of 24.8 Miles...

The EPA Fuel Economy Estimates for automobile models tested recently predicted a mean of 24.8 Miles per gallon, with a standard deviation of 6.2 miles per gallon. Assume that a normal model applies. Find the probability that a randomly selected automobile will average

1 a) Less than 20 Miles per gallon

1 b) More than 14 Miles per gallon

1c ) Between 25 and 30 miles per gallon

Solution :

Given that ,

mean = = 24.8

standard deviation = = 6.2

a) P(x < 20) = P[(x - ) / < (20 - 24.8) /6.2 ]

= P(z < -0.77)

= 0.2206

Probability = 0.2206

b) P(x > 14) = 1 - P(x < 14)

= 1 - P[(x - ) / < (14 - 24.8) /6.2 )

= 1 - P(z < -1.74)

= 1 - 0.0409 = 0.9591

Probability = 0.9591

c) P(25 < x < 30) = P[(25 - 24.8)/ 6.2 ) < (x - ) /  < (30 - 24.8) / 6.2) ]

= P(0.03 < z < 0.84)

= P(z < 0.84) - P(z < 0.03)

= 0.7995 - 0.512 = 0.2875

Probability = 0.2875

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