Q 17RP. - Elementary Statistics

Question

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.

Step-by-Step Solution

Answer

Part (a) $P(\overline{X}>7.5)=5.82\%$

Part (b) No.

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

$$\begin{gathered} {\mu }_x={\mu }_x \\ =6.83 \end{gathered}$$

Sampling distribution of the sample standard deviation

$$\begin{gathered} {\sigma }_{\overline{x}}=\frac{{\sigma }_x}{\sqrt{n}} \\ =\frac{4.28}{\sqrt{100}} \\ =0.428 \end{gathered}$$

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

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

$$\begin{gathered} =P(z>1.57) \\ =1-P(Z\le 1.57) \\ =1-0.9418 \\ =0.0582 \\ =5.82\% \end{gathered}$$

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).