An automobile manufacturer produces a certain model of car. The fuel economy fig

Question

An automobile manufacturer produces a certain model of car. The fuel economy figures of these cars are normally distributed with a mean mileage per gallon (mpg) of 36.8 and a standard deviation of 1.3.

(a) What is the probability that one of these cars will have an mpg of more than 37.5?

(b) What is the probability that one of these cars will have an mpg of less than 35? l

(c) What is the probability that one of these cars will have an mpg of less than 40?

(d) What is the probability that one of these cars will have an mpg of between 34 and 38 mpg?

(e) What is the minimum mpg of the 15% most fuel efficient cars?

(f) What is the maximum mpg of the 10% least fuel efficient cars

Explanation / Answer

Solution:
Given in the question
Mean. =36.8
Standard deviaition = 1.3
Solution(A)
P(Xbar>37.5)=?
Z = (37.5-36.8)/1.3 = 0.5384
So from z-table we found that
P(Xbar>37.5) = 1-P(Xbar<37.5) = 1-0.7054 = 0.2946
so there is 29.46% chances of cars having an mpg more than 37.5
Solution(b)
P(Xbar<35)
Z = (35-36.8)/1.3 = -1.3846
So p-value = 0.0838 or 8.38%
So there is 8.38% chances that these cars will have an mpg of less tha 35
Solution(C)
P(Xbar<40)=?
Z = (40-36.8)/1.3 = 2.4615
P_value = 0.9931
so there is 99.31% chances that cars will have an mpg of less than 40.
Solution(d)
P(34<xbar<38) = P(xbar<38)-P(Xbar<34)
Z = (38-36.8)/1.3 = 0.9230
Z = (34-36.8)/1.3 = -2.1538
So
P(34<xbar<38) = 0.8212- 0.0158 = 0.8054
so there is 80.54% probability that cars will have an mpg of between 34 and 38mpg

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