An automobile manufacturer introduces a new model that has an advertised mean in

Question

An automobile manufacturer introduces a new model that has an advertised mean in-city mileage of 27 miles per gallon and the standard deviation of 3 miles per gallon. Let the random variable X be in-city mileage and assume that it has a normal distribution. What is the probability that a car of this model has less than 22 miles per gallon for in-city driving? Find a value x such that 25% of cars of this model have in-city mileage greater than x miles. Within what limits (centered about the mean) would you expect that X to fall 88% of time? We select 25 cars at random. What is the probability that the average in-city mileage of these 25 cars will be less than 28.5 miles per gallon? What is the probability that the average in-city mileage of these 25 cars will be between 26.4 to 28.5 miles per gallon?

Explanation / Answer

(a)For X=22, z=(x-mu)/sigma

=(22-27)/3=-1.67

P(X<22)=P(z<-1.67)=0.0475

(b)For the probability 0.25, the corresponding z score is: -0.69

Thus, the required raw score X for which 25% of this model have in city mileage greater than X miles is:

X=mu+z*sigma=27-0.69*3=24.93

(d)using Central limit theorem the sampling distribution of sample mean is normally distributed with mean mu and standard deviation sigma/root over n.

So, if 25 cars are selected at random then mean is same as population mean 27 and standard deviation 3/root over 25=0.6

For X bar=28.5, z=(28.5-27)/0.6=2.5

P(X bar<28.5)=P(z<2.5)=0.9938

(e)For X bar=26.4, z=(26.4-27)/0.6=-1

P(26.4<X bar<28.5)=P(X bar<28.5)-P(X bar<26.4)=P(z<2.5)-P(z<-1)

=0.9938-0.1587=0.8351

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