Question

The distribution of actual weights of 8-oz chocolate bars produced by a certain machine is normal with mean 8.3 ounces and standard deviation 0.19 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 8.2 and 8.49 ounces?
ANSWER:

(b) For a SRS of four of these chocolate bars, what is the level LL such that there is a 5% chance that the average weight is less than LL?
ANSWER:

Answer 1

Solution :

Given that ,

mean = = 8.3

standard deviation = = 0.19

P(8.2 < x < 8.49) = P[(8.2 - 8.3)/ 0.19) < (x - ) /  < (8.49 - 8.3) / 0.19) ]

= P(-0.53 < z < 1)

= P(z < 1) - P(z < -0.53)

= 0.8413 - 0.5961

= 0.2452

Probability = 0.2452

b)

P(Z < -1.645) = 0.05

z = -1.645

Using z-score formula,

x = z * +

x = -1.645 * 0.19 + 8.3 = 7.99

L = 7.99