Question

The distribution of actual weights of 80 ounces wedges of cheddar Cheese produced at a dairy is Normal with mean 8.1 ounce and standard deviation is 0.1 ounce. If there is only a 5% chance that the average weight of the sample of five of the cheese wedges will be below ___________ ( get the sample mean value to 2 decimals)

z score:    Answer format: #.##

Average Weight:    Answer format: #.##

Answer 1

Solution :

Given that ,

mean = = 8.1

standard deviation = = 0.1

n = 80

= = 8.1

= / n = 0.1 / 80 = 0.01118

The z - distribution of the 5% is,

P( Z < z ) = 5%

P( Z < z ) = 0.05

P( Z < -1.65 ) = 0.05

z = -1.65

Using z - score formula,

= z * +

= -1.65 * 0.01118 + 8.1

= 8.08

There is only a 5% chance that the average weight of the sample of five of the cheese wedges will be below 8.08.