Question

A brewery's filling machine is adjusted to fill bottles with a mean of 30.9 oz. of ale and a variance of 0.003. Periodically, a bottle is checked and the amount of ale noted.

(a) Assuming the amount of fill is normally distributed, what is the probability that the next randomly checked bottle contains more than 30.94 oz? (Give your answer correct to four decimal places.)


(b) Let's say you buy 92 bottles of this ale for a party. How many bottles would you expect to find containing more than 30.94 oz. of ale? (Round your answer up to the nearest whole number.)
bottles

Answer 1

Solution :

Given ,

mean = = 30.9

variance = 2=0.003

standard deviation = = 0.055

P(x >30.94 ) = 1 - P(x< 30.94)

= 1 - P[(x -) / < (30.94-30.9) /0.055 ]

= 1 - P(z <0.73 )

Using z table

= 1 - 0.7673

= 0.2327

probability= 0.2327

b.

n=92

P(x >30.94 ) = 1 - P(x< 30.94)

= 1 - P[(x -) / < (30.94-30.9) /0.055 ]

= 1 - P(z <0.73 )

Using z table

= 1 - 0.7673

= 0.2327*n

=0.2327*92

=21.4084

=21 rounded