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 value in which 90% of the six-packs will have a higher average content?

Answer 1

Given that,

mean = = 300

standard deviation = = 5

n = 6

= 300

= / n = 5/ 6=2.0412

Using standard normal table,

P(Z > z) = 90%

= 1 - P(Z < z) = 0.90

= P(Z < z ) = 1 - 0.90

= P(Z < z ) = 0.90

= P(Z < -1.28 ) = 0.10

z = -1.28

Using z-score formula  

= z * +

= -1.28 *2.0412+300

= 297.39