Question

In a recent year, about two-thirds of U.S. households purchased ground coffee. Consider the annual ground coffee expenditures for households purchasing ground coffee, assuming that these expenditures are approximately distributed as a normal random variable with a mean of $75 and a standard deviation of $10.

a. Find the probability that a household spent less than $65.00.

b. Find the probability that a household spent more than $80.00.

c. What proportion of the households spent between $65.00 and $80.00?

d. 99% of the households spent less than what amount?

Show all work.

Answer 1

Mean = = 75

Standard deviation = = 10

a)

We have to the probability that a household spent less than $65.00.

That is we have to find P(X < 65)

For finding this probability we have to find z score.

That is we have to find P(Z < - 1)

P(Z < - 1) = 0.1587    ( Using z table)

b)

We have to find the probability that a household spent more than $80.00.

That is we have to find P(X > 80)

For finding this probability we have to find z score.

That is we have to find P(Z > 0.5)

P(Z > 0.5) = 1 - P(Z < 0.5) = 1 - 0.6915 = 0.3085

c)

We have to find the proportion of the households spent between $65.00 and $80.00

That is we have to find P( 65 < X < 80)

For finding this probability we have to find z score.

That is we have to find P( - 1 < Z < 0.5)

P( - 1 < Z < 0.5) = P(Z < 0.5) - P(Z < - 1) = 0.3085 - 0.1587 = 0.1499

d)

We have given P(X < x) = 0.99

z value 0.99 is 2.33

We have to find value of x

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