Suppose the the caffeine content in the cans of a particular brand of cola follo

Question

Suppose the the caffeine content in the cans of a particular brand of cola follows a Normal Model witha mean of 44.5 mg and a standard deviation of 2.95 mg.

a. What percentage of the cans have at most 50 mg of caffeine

b. Ten percent of the cans have more than how many mg of caffeine?

c. In a carton to twenty-four randomly selveted cans of cola, what is the probability that the mean caffeien content for the carton is between 40 and 45 mg?

d. Eighty percent of the cartons of twenty-four randomly selcted cans have a mean caffeine content less than how many mg?

Explanation / Answer

a)percentage of the cans have at most 50 mg of caffeine=P(X<50)=P(Z<(50-44.5)/2.95)=P(Z<1.8644)=0.9689~96.89%

b)for top 10 percentile ; z=1.28

therefore corresponding caffeine amount =mean +z*std deviation =48.28 mg

c)for n=24 ; std error of mean =std deviation/(n)1/2 =0.602

therefore probability that the mean caffeien content for the carton is between 40 and 45 mg

=P(40<X<45)=P((40-44.5)/0.602<Z<(45-44.5)/0.602)=P(-7.473<Z<0.8303)=0.7968-0.0000=0.7968

d) for 80 percentile ; z=0.8416

therefore corresponding mean caffeine content =mean +z*std deviation =44.5+0.8416*0.602=45.01mg

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