Question

The contents of bottles of beer are Normally distributed with a mean of 300 ml and a standard deviation of 5 ml.

What is the probability that the average contents of a six-pack will be between 293 ml and 307 ml?

Answer 1

Solution :

Given that,

mean = = 300

standard deviation = = 5

n=6

= 300

=  / n = 5/ 6=2.04

= P(293<    <307 ) = P[(293-300) / 2.04< ( - ) / < (307-300) / 2.04)]

= P( -3.43< Z <3.43 )

= P(Z <3.43 ) - P(Z < -3.43)

Using z table,  

= 0.9997-0.0003

=0.9994

=