A new car that is a gas- and electric-powered hybrid has recently hit the market. The distance travelled on 1 gallon of fuel is normally distributed with a mean of 60 miles and a standard deviation of 6 miles. Find the probability of the following events:
A. The car travels more than 63 miles per gallon.
Probability =
B. The car travels less than 52 miles per gallon.
Probability =
C. The car travels between 52 and 66 miles per gallon.
Probability =
Solution :
A.
P(x > 63) = 1 - P(x < 63)
= 1 - P[(x - ) / < (63 - 60) / 6)
= 1 - P(z < 0.5)
= 1 - 0.6915
= 0.3085
Probability = 0.3085
B.
P(x < 52) = P[(x - ) / < (52 - 60) / 6]
= P(z < -1.33)
= 0.0918
Probability = 0.0918
C.
P(52 < x < 66) = P[(52 - 60)/ 6) < (x - ) / < (66 - 60) / 6) ]
= P(-1.33 < z < 1)
= P(z < 1) - P(z < -1.33)
= 0.8413 - 0.0918
= 0.7495
Probability = 0.7495
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