Question

The distribution of actual weights of 8-oz chocolate bars produced by a certain machine is normal with mean 7.7 ounces and standard deviation 0.15 ounces.

(a) What is the probability that the average weight of a bar in a Simple Random Sample (SRS) with four of these chocolate bars is between 7.59 and 7.86 ounces?

(b) For a SRS of four of these chocolate bars,

c) what is the level L such that there is a 3% chance that the average weight is less than L?

Answer 1

Given,

= 7.7 , = 0.15

a)

Using central limit theorem,

P( < x) = P( Z < x - / ( / sqrt(n) ) )

So,

P(7.59 < < 7.86) = P( < 7.86) - P( Z < 7.59)

= P( Z < 7.86 - 7.7 / 0.15 / sqrt(4) ) - P( Z < 7.59 - 7.7 / 0.15 / sqrt(4) )

= P( Z < 2.1333) - P( Z < -1.4667)

= 0.9835 - 0.0712

= 0.9123

b)

We have to calculate L such that

P( < L ) = 0.03

That is

P(  Z < L - 7.7 / ( 0.15 / sqrt(4) ) ) = 0.03

From the Z table, z-score for the probability of 0.03 is -1.8808

So,

L - 7.7 / ( 0.15 / sqrt(4) ) = -1.8808

Solve for L

L - 7.7 = -1.8808 * (0.15 / sqrt(4) )

L = 7.5589