Question

Question # 1:A dispensing machine is set to produce 1 pound lots of a particular compound. The machine is fairly accurate, producing mean weight of lots equal to 1.0 pound with a standard deviation of 0.12 pounds. Thirty five lots are randomly selected:

(a)What is the mean weight of the sample mean of the thirty five randomly selected lots?

(b)What is the standard deviation of weight of the sample mean of the thirty five randomly selected lots?

(c)Find the probability that the mean weight is greater than 1.1 pounds.

(d)Find the probability that the mean weight is less than 0.95 pounds

Answer 1

(a)

= Population mean = 1

= Population SD = 0.12

n = Sample size = 35

So,

by Central Limit Theorem:

Mean weight of sample mean = Population mean = 1

So,

Answer is:

1

(b)

Standard deviation of sample mean = SE = /

                                                           = 0.12/ = 0.0203

So,

Answer is:

0.0203

(c)

To find P(>1.1):
Z = (1.1 - 1)/0.0203 = 4.93

Table of Area Under Standard Normal Curve gives area = 0.49999959

So,

P(>1.1) = 0.5 - 0.49999959 = 0.00000041

So,

Answer is:

0.00000041

(d)

To find P(<0.95):
Z = (0.95 - 1)/0.0203 = 2.4631

Table of Area Under Standard Normal Curve gives area = 0.4931

So,

P(<0.95) = 0.5 - 0.4931 = 0.0069

So,

Answer is:

0.0069