315,756 views
16 votes
16 votes
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.

User Gordon Henriksen
by
2.8k points

2 Answers

16 votes
16 votes

Answer:

Mean: 70

Standard Deviation: 4.1

Explanation:

Khan Academy Answer

User Taudorf
by
3.3k points
21 votes
21 votes

Answer:

The answer is below

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

User Bryce Kahle
by
2.8k points