Suppose the national average for a gallon of premium gasoline is $2.65 with a st

Question

Suppose the national average for a gallon of premium gasoline is $2.65 with a standard deviation of 50 cents. Suppose we increase the size of the sample to 100. If the price of a gallon of gasoline follows a normal distribution, compute the results for the sample mean.

What is the probability that the sample mean is at least 2.70 a gallon?

What is the probability that the sample mean is no more than 2.68 a gallon?

What is the probability that the sample mean is between 2.63 and 2.67 a gallon?

What is the probability that the sample mean is within 5 cents of the population mean?

What is the probability that the sample mean is within 10 cents of the population mean?

Explanation / Answer

THE MEAN = 2.65

STANDARD DEVIATION = 0.5

N = 100

DISTRIBUTION TO BE NORMAL AS N >30

FOR MULA TO BE USED = Z = (X-MEAN)/(STANDARD DEVIATION/SQRT(N))

A) P(X>2.65) =

For x = 2.70, z = (2.70 -2.65 / (0.5/SQRT(100)) = 1

Hence P(x > 2.70) = P(z > 1) = [total area] - [area to the left of 1]

1 - [area to the left of 1]

now from the z table we will take the value of z score = 1

    = 1 - 0.8413 = 0.1587

B) P(X<2.68) =

For x = 2.68, z = (2.68 -2.65) / (0.5/SQRT(100)) = 0.6

Hence P(x < 2.68) = P(z < 0.6) =   

[area to the left of 0.6]

now from the z table we will take the value of z score = 0.6

    = 0.7257

C) P(2.63<X<2.67) =

For x = 2.63 , z = (2.63 - 2.65) /(0.5/SQRT(100) = -0.40

and for x = 2.67, z = (2.67 - 2.65) /(0.5/SQRT(100) = 0.4

Hence P(2.63 < x < 2.67) = P(-0.4 < z < 0.4) = [area to the left of z = 0.4] - [area to the left of -0.4]

= 0.6554 - 0.3446 = 0.3108

D) WITHIN 5 CENTS = P(2.60<X<2.70) =

For x = 2.60 , z = (2.60 - 2.65) /(0.5/SQRT(100) = -1

and for x = 2.67, z = (2.7 - 2.65) /(0.5/SQRT(100) = 1

Hence P(2.60 < x < 2.7) = P(-1 < z <1) = [area to the left of z = 1] - [area to the left of -1]

= 0.8413 - 0.1587 = 0.6826

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