Shoppers at a mall have a mean weight of 70 kg 70kg70, start text, k, g, end text with a standard deviation of 10 kg 10kg10, start text, k, g, end text. An elevator at the mall holds a maximum of 6 66 people, and safety engineers are curious about the average weight of shoppers on a full elevator. Suppose that we take random samples of 6 66 shoppers and calculate the mean weight x ˉ x ˉ x, with, \bar, on top of the shoppers in each sample.

The answer is below

Step-by-step explanation:

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 x ˉ  on top of the shoppers in each sample.

Solution:

Let variable x represent the weight of a shopper at the mall.

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

There are random samples of 6 shoppers. That is sample size (n) = 6  

The mean of the sample (μₓ) is the same as the mean of the population (μ), hence:

μₓ = μ = 70 kg

The standard deviation of the sample (σₓ) is equal to the standard deviation of the population (σ) divided by the square root of the sample size (n).. Hence:

σₓ = σ / √n = 10 / √6 = 4.08 kg

Mean: 70

Standard Deviation: 4.1

Step-by-step explanation:

Khan Academy Answer

option 1 is correct.

step-by-step explanation:

y varies inversely with x and constant of variation is 4

the equation will be:

y = k/x

where k =4 so,

y = 4/x

if we put values of x of option 1 in the above equation and find values of y

y = 4/-2 => y =-2

y = 4/-1 => y = -4

y = 4/1 => y = 4

y = 4/2 => y =2

so, all the values of x and y are correct.

if we put values of x given in other options and find value of y we cannot find whole table satisfying the equation.

hence option 1 is correct.