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Found in: Page 311

### Elementary Statistics

Book edition 9th
Author(s) Weiss, Neil
Pages 590 pages
ISBN 9780321989505

# 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) $P\left(\overline{X}>7.5\right)=5.82%$

Part (b) No.

See the step by step solution

## 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 ${\mu }_{x}=6.83$ and ${\sigma }_{x}=4.28$

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

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

## Part (a) Step 2: Concept

The formula used: Standard deviation ${\sigma }_{\overline{x}}=\frac{{\sigma }_{x}}{\sqrt{n}}$

## 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

${\mu }_{x}={\mu }_{x}\phantom{\rule{0ex}{0ex}}=6.83$

Sampling distribution of the sample standard deviation

${\sigma }_{\overline{x}}=\frac{{\sigma }_{x}}{\sqrt{n}}\phantom{\rule{0ex}{0ex}}=\frac{4.28}{\sqrt{100}}\phantom{\rule{0ex}{0ex}}=0.428$

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

That is $P\left(\overline{X}>7.5\right)$

$=P\left(z>1.57\right)\phantom{\rule{0ex}{0ex}}=1-P\left(Z\le 1.57\right)\phantom{\rule{0ex}{0ex}}=1-0.9418\phantom{\rule{0ex}{0ex}}=0.0582\phantom{\rule{0ex}{0ex}}=5.82%$

## 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).