Question

The company manufactures chocolate bars with nominal weight of 115 g per bar. In order to to avoid risk of short changing, the mean weight is set as 117 g and SD of 2 g.

Assuming that 40 chocolate bars are selected, 97% of the sample means will be between what value? ( give the upper and lower value)

Answer 1

Solution:-

Given that,

mean = = 117

standard deviation = = 2

n = 40

= = 117

= / n = 2 / 40 = 0.316

Using standard normal table,

P( -z < Z < z) = 97%

= P(Z < z) - P(Z <-z ) = 0.97

= 2P(Z < z) - 1 = 0.97

= 2P(Z < z) = 1 + 0.97

= P(Z < z) = 1.97 / 2

= P(Z < z) = 0.985

= P(Z < 2.17) = 0.985

= z  ± 2.17

Using z-score formula  

= z * +   

= -2.17 * 0.316 + 117

= 116.3

Using z-score formula  

= z * +   

= 2.17 * 0.316 + 117

= 117.7

A 97% value = ( 116.3, 117.7 )