Question

In: Statistics and Probability

The annual per capita (average per person) chewing gum consumption in the United States is 200...

The annual per capita (average per person) chewing gum consumption in the United States is 200 pieces. Suppose that the standard deviation of per capita consumption of chewing gum is 145 pieces per year.

(a) Find the probability that the average annual chewing gum consumption of 84 randomly selected Americans is more than 220 pieces.

(b) Find the probability that the average annual chewing gum consumption of 84 randomly selected Americans is within 100 pieces of the population mean.

(c) Find the probability that the average annual chewing gum consumption of 16 randomly selected Americans is less than 100 pieces.

Solutions

Expert Solution

P(z<Z) table :

a.

SD of sample = (population SD)/(n^0.5)

= 145/(84^0.5)

= 15.82

b.

SD of sample = (population SD)/(n^0.5)

= 145/(84^0.5)

= 15.82

c.

SD of sample = (population SD)/(n^0.5)

= 145/(16^0.5)

= 36.25

(please UPVOTE)


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