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frosja888 [35]
2 years ago
9

A study on the latest fad diet claimed that the amounts of weight lost by all people on this diet had a mean of 23.4 pounds and

a standard deviation of 6.8 pounds.
Step 2 of 2 : If a sampling distribution is created using samples of the amounts of weight lost by 63 people on this diet, what would be the standard deviation of the sampling distribution of sample means? Round to two decimal places, if necessary.
Mathematics
1 answer:
ad-work [718]2 years ago
4 0

Using the Central Limit Theorem, the standard deviation of the sampling distribution of sample means would be of 0.86.

<h3>What does the Central Limit Theorem state?</h3>

It states that the standard deviation of the sampling distribution of sample means is given by:

s = \frac{\sigma}{\sqrt{n}}

In which:

  • \sigma is the standard deviation of the population.
  • n is the sample size.

The parameters for this problem are given as follows:

\sigma = 6.8, n = 63.

Hence:

s = \frac{\sigma}{\sqrt{n}}

s = \frac{6.8}{\sqrt{63}}

s = 0.86.

More can be learned about the Central Limit Theorem at brainly.com/question/16695444

#SPJ1

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Answer:

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Step-by-step explanation:

Given:

Width = 0.4

Let's take the z value of 95% which is = 1.96

Let's assume population proportion p1 and p2 = 0.5

For margin of error E, the relation between the width and margin of error is 2E.

i.e 2E = 0.4

E = 0.2

To find the sample size, n, let's use the formula :

n= \frac{z^2*(p_1(1-p_1) + p_2(1-p_2))}{E^2}

= \frac{1.96^2(0.5(1-0.5)+0.5(1-0.5))}{0.2^2}

= 48.02

≈ 49

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Answer:

Area = length x width

width = 1/2 length = 12/2 = 6 = width

area = 6x 12 = 72

Hope that answers your question

don't hesitate to comment if you are confused about something

Step-by-step explanation:

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