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.

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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

(1,4) is on graph of f(x)

f(1) = 4

so 1/4 f(x)

1/4 f(1)

1/4. ×4

1

so point for function 1/4. f(x) is (1,1)

i think 5 miles per hour

step-by-step explanation: