Question

The number of wiring packages that can be assembled by a company's employees has a normal distribution, with a mean equal to 19.8 per hour and a standard deviation of 1.2 per hour.

(a) What are the mean and standard deviation of the number X of packages produced per worker in an 8-hour day? (Round your standard deviation to three decimal places.)
a. mean
b. standard deviation
(b) Do you expect the probability distribution for X to be mound-shaped and approximately normal? Explain.
Choose
A. Yes, since the standard deviation is less than 3, the sampling distribution of the sum will be approximately normal.
B. Yes, since the original population is normal, the sampling distribution of the sum will also be approximately normal.
C. No, since the original population is not normal, the sampling distribution of the sum will not be approximately normal.
D. No, since the original population is normal, the sampling distribution of the sum cannot be normal.
E. No, since the standard deviation is more than 3, the sampling distribution of the sum will not be approximately normal.
(c) What is the probability that a worker will produce at least 160 packages per 8-hour day? (Round your answer to four decimal places.)

Answer 1

a) mean= 158.4 , standard deviation = 3.394

b) Best option : B. Yes, since the original population is normal, the sampling distribution of the sum will also be approximately normal.

c) P(X>160) = P(Z>0.471) = 1-P(Z<0.471) = 0.3188

Explanation:

1) Notation

n = sample size = 8

`\mu` = population mean = 19.8

`\sigma` = population standard deviation = 1.2

2) Definition of the variable of interest

Part a

The variable that we are interested is
`\sum x_i` and the mean and the deviation for this variable are given by :

E(
`\sum x_i`) =
`\sum E(x_i)` = n
`\mu` = 8x19.8 = 158.4

Var(
`\sum x_i`) =
`\sum Var(x_i)` = n
`\sigma^2`

Sd(
`\sum x_i`) =
`√(n \sigma^2)` =
`\sqrt(8)` x 1.2 = 3.394

Part b

For this case the populations are normal, then the distribution for the sample (
`\sum x_i`) is normal too.

Based on this the distribution for the variable X would be normal, so the best option should be:

B. Yes, since the original population is normal, the sampling distribution of the sum will also be approximately normal.

Part c

From part a we know that the mean = 158.4 and the deviation = 3.394

The z score is defined as

Z = (X -mean)/ deviation = (160-158.4)/ 3.394 = 0.471

Then we can find the probability P(X>160) = P(Z>0.471) = 1-P(Z<0.471) = 0.3188