Shoppers at a mall have a mean weight of 70 kg with a standard deviation of 10 kg. An elevator at the mall holds a maximum of 6 people, and safety engineers are curious about the average weight of shoppers on a full elevator. Suppose that we take random samples of 6 shoppers and calculate the mean weight of the shoppers in each sample.

If there are random samples of 6 shoppers taken from a normal population, the sample mean of those samples will be a random variable with E(X)= μ and V(X)= σ²/n

X~N(μ;σ²/n)

The population mean of the sampling distribution is

E(X)= μ= 70 Kg

The population variance and standard deviation of the sampling distribution are

V(X)= σ²/n= 10²/6= 16.67

√V(X)= 4.08

I hope it helps!

keshavgandhi04 keshavgandhi04 · Answer 2 · 2021-12-17T16:44:36+01:00

For the sampling distribution, the standard deviation is 4.08 and for the sampling distribution, the formula of the population variance is 16.67 and this can be determined by using the given data.

Given :

Shoppers at a mall have a mean weight of 70 kg with a standard deviation of 10 kg.
An elevator at the mall holds a maximum of 6 people, and safety engineers are curious about the average weight of shoppers on a full elevator.

For the sampling distribution the population mean is given below:

[tex]\rm E(x) = \mu = 70\; Kg[/tex]

Now, for the sampling distribution, the formula of the population variance is given below:

[tex]\rm V(x) = \dfrac{\sigma^2}{n}[/tex]

Now, substitute the values of the known terms in the above formula.

[tex]\rm V(x) = \dfrac{(10)^2}{6}[/tex]

V(x) = 16.67

Now, for the sampling distribution, the formula of the standard deviation is given below:

[tex]\rm \sqrt{V(x)} = \sqrt{\dfrac{10^2}{6}} =4.08[/tex]

For more information, refer to the link given below:

https://brainly.com/question/12402189