Find the probability of a household


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Q: In recent years, about two-thirds of US house holds 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 $65.16 and a standard deviation of $10.00

Using the normal probability:

A. Find the probability that a household spent less than $35.00.

B. Find the probability that a household spent more than $60.00.

C. What proportion of housholds spent between $40.00 and $50.00.

D. 99% of housholds spent less than what amount?

Normal Probability

Mean = 65.16

Standard Deviation = 10

Compute probability: X < 35 ==> 0.13%

Compute probability: X > 60 ==> 69.71%

Compute probability: Range 40 <= X <= 450 ==> 99.41% (i.e., between 2 values)

Find actual score: Percentage ABOVE = 5.00% ==> 81.61 (i.e. % greater than)

Find actual score: Percentage BELOW = 95.00% ==> 81.61 (i.e. % less than)

985_freq.jpg

Approximately 68% of all observations within 1 standard deviation of the mean

Approximately 95% of all observations within 2 standard deviations of the mean

Approximately 99.7% of all observations within 3 standard deviations of the mean

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Basic Statistics: Find the probability of a household
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