Question

Suppose that the mean retail price per gallon of regular grade gasoline in the United States is $3.45 with a standard deviation of $0.20 and that the retail price per gallon has a bell-shaped distribution. (a) What percentage of regular grade gasoline sold between $3.25 and $3.65 per gallon? % (b) What percentage of regular grade gasoline sold between $3.25 and $3.85 per gallon? % (c) What percentage of regular grade gasoline sold for more than $3.85 per gallon? %

Answer 1

The retail price has a bell-shaped distribution which corresponds to a standard normal distribution and hence we will be using a standard normal probability table.

Given

Mean = $ 3.45

Standard deviation = $ 0.20

a) 68 .26 %

Below is the working

We need to find the z score for $ 3.25 & $ 3.65 and based on that we need to find probabilities

so z score = ( x - ) /

so at x = $ 3.25

z score = ( 3.25 - 3.45 ) / 0.20

z score = - 1

Now at x = 3.65

z score = (3.65 - 3.45) / 0.2

z score = 1

from question we need to find

P( -1 < z < 1) = P( z < 1) - P (z < -1)

From standard normal distribution table

P( z < 1) = 0.84134

P (z < -1) = 0.15866

substituting above we get

P ( -1 < z < 1) = 0.84134 - 0.15866

P ( -1 < z < 1) = 0.682688 = 68.26 %

so 68.26 % of regular gasoline sold b/w $ 3.25 & $ 3.65

b) 81.85 %

Below is the working

We need to find the z score for $ 3.25 & $ 3.85 and based on that we need to find probabilities

so z score = ( x - ) /

so at x = $ 3.25

z score = ( 3.25 - 3.45 ) / 0.20

z score = - 1

Now at x = 3.85

z score = (3.85 - 3.45) / 0.2

z score = 2

from question we need to find

P( -1 < z < 2) = P( z < 2) - P (z < -1)

From standard normal distribution table

P( z < 2) = 0.97725

P (z < -1) = 0.15866

substituting above we get

P ( -1 < z < 1) = 0.97725 - 0.15866

P ( -1 < z < 1) = 0.81859 = 81.85 %

so 81.85 % of regular gasoline sold b/w $ 3.25 & $ 3.85

c) 2.27 %

We need to find the z score for $ 3.85 and based on that we need to find probabilities

so z score = ( x - ) /

Now at x = 3.85

z score = (3.85 - 3.45) / 0.2

z score = 2

from question we need to find -

P( z > 2) = 1 - P (z < 2) ----- (Note Probability of a sure activity is 1)

From standard normal distribution table

P( z < 2) = 0.97725

substituting above we get

P (z > 2) = 1 - 0.97725

P ( z > 2 ) = 0.02275 = 2.27 %

so 2.27 % of regular gasoline sold for more than $ 3.85

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