what is the probability?

Question

During 2001, 61.3% of US households purchased ground coffee and spent an average of $36.16 on ground coffee during the year (annual product preference study) ground coffee expenditures for households purchasing approximately distributed as a normal random variable with a mean of $36.16 and a standard deviation of $10.00. used the standard normal cumulative probability table

a. Find the probability that a household spent less than $25.00.
b. Find the probability that a household spent more than $50.00?
c. What the proportion of the households spent between 30.00 and $40.00?
d. 99% of the households spent less than what amount?

Answer 1

For any normal random variable X with mean μ and standard deviation σ, X ~ Normal(μ, σ)

you can translate into standard normal units by:
Z = (X - μ) / σ

where Z ~ Normal(μ = 0, σ = 1). You can then use the standard normal cdf tables to get probabilities.

Let X be the amount spent on coffee by those who purchase coffee

X ~ Normal(μ = 36.16, σ = 10.00)

a) P( X < 25)
= P(Z < (25 - 36.16) / 10)
= P( Z < -1.116)
= 0.1322111

b) P(X > 50)
= P(Z > (50 - 36.16) / 10)
= P( Z > 1.384)
= 0.08317923

c) P(30 < X < 40)
= P(X < 40) - P(X < 30)
= 0.6495108 - 0.2689473
= 0.3805635

d) P(Z < z) = 0.99
z = 2.326348

X = 36.16 + 2.326348 * 10.00
X = 59.42348

99% of the people who by coffee spend less than $59.42348

Answer 2

Translate each of the numbers $25, $50, $30, and $40 into "So and so many standard deviations above or below the mean"

Then go to town with your table. That will take care of a-c.