Suppose a bottle of heinz Ketchup is normally distributed wi

Suppose a bottle of heinz Ketchup is normally distributed with a mean of 16oz. and a Standard Deviation of .12 oz.

First calculate the z-value for this problem.

a. what percent of bottles have less than15.9 oz?

b. What pecrent of bottles have more than15.9 oz?

c. What percent of bottles have between 15.9 and 16.1 oz?

Solution

a. what percent of bottles have less than15.9 oz?

P(X<15.9) = P((X-mean)/s <(15.9-16)/0.12)

=P(Z<-0.83) =0.2033 (from standard normal table)

i.e. 20.33%

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b. What pecrent of bottles have more than15.9 oz?

P(X>15.9) = P(Z>-0.83) = 0.7967 (from standard normal table)

i.e. 79.67%

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c. What percent of bottles have between 15.9 and 16.1 oz?

P(15.9<X<16.1) = P((15.9-16)/0.12<Z<(16.1-16)/0.12)

=P(-0.83<Z<0.83) =0.5935 (from standard normal table)

i.e. 59.35%