Jace gathered the data in the table. He found the approximate line of best fit to be y = –0.7x + 2.36.

Mathematics

1 answer:

MaRussiya [10]3 years ago

3 0

The residual value is -1.14.
Plug 5 into x
y=-0.7(5)+2.36
=-1.14

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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, sta

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Answer:

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

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<h3>Answer: choice A) 55.3%</h3>

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Explanation:

Use a Z table found in the back of your book to find that

P(Z < -0.41) = 0.3409

P(Z < 1.25) = 0.8944

So,

P(-0.41 < Z < 1.25) = P(Z < 1.25) - P(Z < -0.41)

P(-0.41 < Z < 1.25) = 0.8944 - 0.3409

P(-0.41 < Z < 1.25) = 0.5535

Now convert this to a percentage by multiplying by 100, which is the same as moving the decimal point over 2 places

0.5535 ---> 55.35%

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