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.

Step-by-step explanation:

Hello!

The variable of interest is X: the weight of a shopper at the mall.

Assuming this variable has a normal distribution with mean μ= 70kg and standard deviation δ=10kg.

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!

the equivalent expression are always the hardest ones but you look at the gcf also

step-by-step explanation:

answer: bearing in mind that a rectangle has a perimeter of length + length + width + width.

step-by-step explanation:

p = perimeter

l = length

x=width

p = 2l + 2x

p = 2(l+x)     < ==== perimeter of the base of the box.