91影视

Given in the question that,

population mean$\left(渭\right)=6$

population standard deviation$\left(蟽\right)=2.4$

sample size $\left(n\right)=10$

we have to find the mean and standard deviation of $\stackrel{炉}{x}$.

Formula used:

${渭}_{\stackrel{炉}{x}}=渭$

${蟽}_{\overline{x}}=\frac{蟽}{\sqrt{n}}$

As per central limit theorm,

$\overline{X}~N\left(渭,\frac{{蟽}^2}{n}\right)$

The mean and standard deviation are:

${渭}_{\overline{x}}=6$

${蟽}_{\stackrel{炉}{x}}=\frac{2.4}{\sqrt{10}}=0.75895$

Thus the mean and standard deviation are$6and0.75895$

Given in the question that the probability that $\stackrel{炉}{x}<5$based on a sample of size $10$we have to explain why it can't safely calculate the probability.

In order to use Central Limit theorem,

The population should follow normality and the sample size should be greater than or equal to $30$. Here, the distribution of population is unknown and the sample size is also less than $30$. Therefore, $\overline{x}$cannot be approximately normal distributed.

Given in the question that Suppose the NLDN takes a random sample of $n=50$square kilometers instead we have to xplain how the central limit theorem allows us to find the probability that the mean number of lightning strikes per square kilometer is less than$5$

The probability can be calculated as follows:

$P(\overline{x}<5)=$$P(\stackrel{炉}{x}<5)=P\left(\frac{{\displaystyle \frac{\overline{x}-渭}{蟽}}}{\sqrt{n}}<\frac{{\displaystyle \frac{5-渭}{蟽}}}{\sqrt{n}}\right)$

$=P\left(Z<\frac{{\displaystyle \frac{5-6}{2.4}}}{\sqrt{10}}\right)$

$=0.0934$

the probability is$0.0934$