Question

(1 point) The distribution of actual weights of 8-oz chocolate bars produced by a certain machine is normal with mean 8 ounces and standard deviation 0.13 ounces.

(a) What is the probability that the average weight of a bar in a Simple Random Sample (SRS) with three of these chocolate bars is between 7.9 and 8.16 ounces? ANSWER:

(b) For a SRS of three of these chocolate bars, what is the level L such that there is a 4% chance that the average weight is less than L? ANSWER:

Answer 1

Solution :

Given that ,

mean = = 8

standard deviation = = 0.13

a) P(7.9 < x < 8.16) = P[(7.9 - 8)/ 0.13) < (x - ) /  < (8.16 - 8 ) / 0.13) ]

= P(-0.77 < z < 1.23)

= P(z < 1.23) - P(z < -0.77)

Using z table,

= 0.8907 - 0.2206

= 0.6701

b) n = 3

= = 8

= / n = 0.13 / 3 = 0.075

Using standard normal table,

P(Z < z) = 4%

= P(Z < z) = 0.04  

= P(Z < -1.75) = 0.04

z = -1.75

Using z-score formula  

= z * +

= -1.75 *0.075 + 8

= 7.87 ounces