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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