Question

Can I have an answer for problem 5.43, page 218 please?

Suppose that a machine filling mini-boxes of raisins fills the boxes so that the weight of the boxes has a population mean = 14.1g, and standard deviation = 1.4g. If a random sample of 49 boxes is selected,

a/ What is the probability that the sample mean will be above 14g?

b/ There is a 99% chance that the sample mean will be above what value?

c/ What assumptions are necessary in order to answer a and b?

d/ If a random sample of 64 boxes is selected, what would be your answers for a and b?

Answer 1

a) P( > 14)

= P(( - )/() > (14 - )/())

= P(Z > (14 - 14.1)/(1.4/))

= P(Z > -0.5)

= 1 - P(Z < -0.5)

= 1 - 0.3085

= 0.6915

b) P(( > x) = 0.99

or, P(( - )/() > (x - )/()) = 0.99

or, P(Z > (x - 14.1)/(1.4/)) = 0.99

or, P(Z < (x - 14.1)/(1.4/)) = 0.01

or, (x - 14.1)/(1.4/) = -2.33

or, x = -2.33 * (1.4/) + 14.1

or, x = 13.634

c) Since the sample size n > 30, so according to the CLT, the sampling distribution of is approximately normally distributed.

d) i) P( > 14)

= P(( - )/() > (14 - )/())

= P(Z > (14 - 14.1)/(1.4/))

= P(Z > -0.57)

= 1 - P(Z < -0.57)

= 1 - 0.2843

= 0.7157

ii) P(( > x) = 0.99

or, P(( - )/() > (x - )/()) = 0.99

or, P(Z > (x - 14.1)/(1.4/)) = 0.99

or, P(Z < (x - 14.1)/(1.4/)) = 0.01

or, (x - 14.1)/(1.4/) = -2.33

or, x = -2.33 * (1.4/) + 14.1

or, x = 13.692