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Q 17RP.

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Elementary Statistics

Found in: Page 311

Book edition 9th

Author(s) Weiss, Neil

Pages 590 pages

ISBN 9780321989505

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Short Answer

Cloudiness in Breslau. In the paper "Cloudiness: Note on a Novel Case of Frequency" (Proceedirgs of the Royal Society of London, Vol. 62. pp. 287-290), K. Pearson examined data on daily degree of cloudiness, on a scale of $0$ to $10$ , at Breslau (Wroclaw), Poland, during the decade $1876 - 1885$ . A frequency distribution of the data is presented in the following table. From the table, we find that the mean degree of cloudiness is $6.83$ with a standard deviation of $4.28$ .
a. Consider simple random samples of $100$ days during the decade in question. Approximately what percentage of such samples have a mean degree of cloudiness exceeding $7.5$ ?
b. Would it be reasonable to use a normal distribution to obtain the percentage required in part (a) for samples of size $5$ ? Explain your answer.

Part (a) $P (\bar{X} > 7.5) = 5.82 %$

Part (b) No.

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Part (a) Step 1: Given information

The frequency distribution of the data is depicted in the table below based on the information provided.

Degree	Frequency	Degree	Frequency
0	751	6	21
1	179	7	71
2	107	8	194
3	69	9	117
4	46	10	2089
5	9

Given the mean degree of cloudiness is $6.83$ with a standard deviation of $4.28$

That is $μ_{x} = 6.83$ and $σ_{x} = 4.28$

Let $X$ denotes the number of degree of cloudiness.

A population variable $x$ has a normal distribution with a mean $μ$ and standard deviation $σ$ The variable $\bar{x}$ is then normally distributed for samples of size $n$ , with a mean $mu$ and standard deviation $σ / \sqrt{n}$

Part (a) Step 2: Concept

The formula used: Standard deviation $σ_{\bar{x}} = \frac{σ_{x}}{\sqrt{n}}$

Part (a) Step 3: Calculation

We need to figure out what percentage of $100 -$ day simple random samples have a mean degree of cloudiness greater than $7.5$

Sample size $n = 100$

Sampling distribution of the sample mean

$μ_{x} = μ_{x} = 6.83$

Sampling distribution of the sample standard deviation

$σ_{\bar{x}} = \frac{σ_{x}}{\sqrt{n}} = \frac{4.28}{\sqrt{100}} = 0.428$

We need to figure out what proportion of the sample mean degree of cloudiness is greater than $7.5$

That is $P (\bar{X} > 7.5)$

$= P (z > 1.57) = 1 - P (Z \leq 1.57) = 1 - 0.9418 = 0.0582 = 5.82 %$

Part (b) Step 1: Explanation

The sample size is $5$ ; the assumption for addressing this problem is that the degree of cloudiness distribution is not normally distributed, as shown in part (A).

Most popular questions for Math Textbooks

A statistic is said to be an unbiased estimator of a parameter if the mean of all its possible values equals the parameter; otherwise, it is said to be a biased estimator. An unbiased estimator yields, on average, the correct value of the parameter, whereas a biased estimator does not.

Part (a): Is the sample mean an unbiased estimator of the population mean? Explain your answer.

Part (b): Is the sample median an unbiased estimator of the population mean? Explain your answer.

What is another name for the standard deviation of the variable $x$ ? What is the reason for that name?

The following graph shows the curve for a normally distributed variable. Superimposed are the curves for the sampling distributions of the sample mean for two different sample sizes.

a. Explain why all three curves are centered at the same place.

b. Which curve corresponds to the larger sample size? Explain your answer.

c. Why is the spread of each curve different?

d. Which of the two sampling-distribution curves corresponds to the sample size that will tend to produce less sampling error? Explain your answer.

c. Why are the two sampling-distribution curves normal curves?

7.43 NBA Champs. Repeat parts (b) and (c) of Exercise 7.41 for samples of size 3. For part (b), use your answer to Exercise 7.13(b).

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