Question

(2 pts) The distribution of actual weights of 8-oz chocolate bars produced by a certain machine is normal with mean 7.8 ounces and standard deviation 0.2 ounces.

(a) What is the probability that the average weight of a random sample of 4 of these chocolate bars will be between 7.66 and 7.9 ounces?
ANSWER:

(b) For a random sample of of these chocolate bars, find the value L such that P(x¯<L)= 0.0281.
ANSWER:

Answer 1

a)

here, mean = 7.8 and standard deviation = 0.2

to find the probability, first, we need to determine the z value

p(7.66<x<7.9) (between 7.66 and 7.9)

so here we need to determine,

p(x<7.66) and p(x<7.9)

p(x<7.66) = 1 - p(x<7.66)

p(x<7.66) = p(x<z)

z = (x-mean)/(standard deviation)

therefore z for 7.66 is (7.66-7.8)/0.2 = -0.7

now from z table -0.7 is 0.2420

therefore, p(x<7.66) = 0.2420

now for p(x<7.9)

z = 7.9-7.8/0.2 = 0.5

from z table, 0.5 is 0.6915

p(7.66<x<7.9) = p(x<7.9) - p(x<7.66)

= 0.6915 - 0.2420

= 0.4495

b)

find the value L such that P(x¯<L)= 0.0281.

in z table -1.91 corresponds to 0.0281

therefore value of l is -1.91