Mathematics: 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,

i know right!?Step-by-step explanation:so annoying...

The number of wiring packages that can be assembled

Karlakins

10.01.2022

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

Step-by-step 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( $\sum x_i$ ) = $\sqrt{n \sigma^2}$ = $\sqrt(8)$ 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

0,0(0 оценок)