Q.60 - The Practice of Statistics for AP

Question

Lightning strikes The number of lightning strikes on a square kilometer of open ground in a year has mean $6$ and standard deviation $2.4$. (These values are typical of much of the (United States.) The National lighting Detection Network (NLDN) uses automatic sensors to watch for lightning in a random sample of 10 one-square-kilometer plots of land.

(a) What are the mean and standard deviation of $\overline{x}$ the sample mean number of strikes per square kilometers?

(b) Explain why you can't safely calculate the probability that $\overline{x}<5$ based on a sample of size $10$ .

(c) Suppose the NLDN takes a random sample of $n=50$ square kilometers instead. Explain how the central limit theorem allows us to find the probability that the mean number of lightning strikes per square kilometer is less than $5$ . Then calculate this probability. Show your work.

Step-by-Step Solution

Answer

(a) Mean $=2.4$

standard deviation $=0.75895$

(b)$\overline{x}$ cannot be approximately normal distributed.

(c) The probability is $0.0934$

Part (a) Step-1 Given Information

Given in the question that,

population mean$\left(\mu \right)=6$

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

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

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

Part (a) Step-2 Explanation

Formula used:

${\mu }_{\overline{x}}=\mu$

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

As per central limit theorm,

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

The mean and standard deviation are:

${\mu }_{\overline{x}}=6$

${\sigma }_{\overline{x}}=\frac{2.4}{\sqrt{10}}=0.75895$

Thus the mean and standard deviation are $6 and 0.75895$

Part (b) Step-1: Given Information

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

Part (b) Step-2 Explanation

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.

Part (c) Step-1 Given Information

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 $5$

Part (c) Step-2: Explanation

The probability can be calculated as follows:

$P(\overline{x}<5)=$$P(\overline{x}<5)=P\left(\frac{{\displaystyle \frac{\overline{x}-\mu }{\sigma }}}{\sqrt{n}}<\frac{{\displaystyle \frac{5-\mu }{\sigma }}}{\sqrt{n}}\right)$

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

                                     $=0.0934$

the probability is $0.0934$