Question

ccording to the U.S. National Weather Service, at any given moment of any day, approximately 1000 thunderstorms are occurring worldwide. Many of these storms include lightning strikes. Sensitive electronic equipment is used to record the number of lightning strikes worldwide every day. 23 days were selected at random, and the number of lightning strikes on each day was recorded. The sample mean was 8.1 million. Assume the distribution of the number of lightning strikes per day is normal and has a population standard deviation of 0.51 million. Please use 4 decimal places for all critical values.

Answer 1

Answer:

The critical value of z for 99% confidence interval is 2.5760.

The 99% confidence interval for population mean number of lightning strike is (7.83 mn, 8.37 mn).

Step-by-step explanation:

Let X = number of lightning strikes on each day.

A random sample of n = 23 days is selected to observe the number of lightning strikes on each day.

The random variable X has a sample mean of, $\bar x=8.1\ mn$ and the population standard deviation, $\sigma=0.51\ mn$.

The (1 - α)% confidence interval for population mean μ is:

$CI=\bar x\pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$

Compute the critical value of z for 99% confidence interval is:

$z_{\alpha/2}=z_{0.01/2}=z_{0.005}=2.5760$

*Use a z-table.

The critical value of z for 99% confidence interval is 2.5760.

Compute the 99% confidence interval for population mean number of lightning strike as follows:

$CI=\bar x\pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}\\=8.1\pm2.5760\times\frac{0.51}{\sqrt{23}}\\=8.1\pm0.2738\\=(7.8262, 8.3738)\\\approx(7.83, 8.37)$

Thus, the 99% confidence interval for population mean number of lightning strike is (7.83 mn, 8.37 mn).

The 99% confidence interval for population mean number of lightning strike  implies that the true mean number of lightning strikes lies in the interval (7.83 mn, 8.37 mn) with 0.99 probability.