91Ó°ÊÓ

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

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

That is ${\mathrm{μ}}_x=6.83$and ${\mathrm{σ}}_x=4.28$

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

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

The formula used: Standard deviation${\mathrm{σ}}_{\stackrel{¯}{x}}=\frac{{\mathrm{σ}}_x}{\sqrt{n}}$

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

$$\begin{gathered} {\mathrm{μ}}_x={\mathrm{μ}}_x \\ =6.83 \end{gathered}$$

Sampling distribution of the sample standard deviation

$$\begin{gathered} {\mathrm{σ}}_{\stackrel{¯}{x}}=\frac{{\mathrm{σ}}_x}{\sqrt{n}} \\ =\frac{4.28}{\sqrt{100}} \\ =0.428 \end{gathered}$$

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

That is $P(\stackrel{¯}{X}>7.5)$

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

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).