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Q 17RP.

Expert-verifiedFound in: Page 311

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

Part (b) No.

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 $\sigma /\sqrt{n}$

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

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 $P(\overline{X}>7.5)$

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

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).

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