Q. 7.51 - Elementary Statistics

Question

7.51 Earthquakes. According to The Earth: Structure, Composition and Evolution (The Open University, S237), for earthquakes with a magnitude of $7.5$ or greater on the Richter scale, the time between successive earthquakes has a mean of $437$days and a standard deviation of $399$ days. Suppose that you observe a sample of four times between successive earthquakes that have a magnitude of $7.5$ or greater on the Richter scale.
a. On average, what would you expect to be the mean of the four times?
b. How much variation would you expect from your answer in part (a)? (Hint: Use the three-standard-deviations rule.)

Step-by-Step Solution

Answer

(a) The mean of the four times in the sample will be equal to $437$ days.

(b) The expected variation from the answer in part (a) is $\pm 598.5$.

Part (a) Step 1: Given information

To find the mean of the four times. Note that the mean is $437$ days and the standard deviation is $399$days.

Part (a) Step 2: Explanation

Earthquakes having a magnitude of $7.5$ or larger on the Richter scale make up the majority of the population.
The mean time between two earthquakes in the population days $\mu =437$.
Population for standard deviation is $399$ days.
Sample size $n=4$

Since,  $\overline{x}$ be the sample mean time between two earthquakes.

So, the mean of $\overline{x}=\mu =437$days.

In other words, because the mean of all possible sample means is equal to population mean $\mu$, will expect the mean of the four times in the sample to be equal to $437$ days on average.
As a result, on an average can expect that mean of the four times in the sample will be equal to $437$ days.

Part (b) Step 3: Given information

To find the expected variation from the answer in part (a) by using the three-standard-deviations rule.

Part (b) Step 4: Explanation

Let, the mean is $=437$ days.
And the standard deviation $=399$ days.
Since the standard deviation of all possible sample mean is ${\sigma }_{\overline{x}}$ as:
$=\frac{\sigma }{\sqrt{n}}$
$=\frac{399}{\sqrt{4}}$$=199.5$days.
Expect $\pm 3{\sigma }_{\overline{x}}$  variation from the mean of $\overline{x}$ which is from population mean $\mu$.
Expected amount of variation from $\overline{x}=\pm 3{\sigma }_{\overline{x}}$
$\begin{array}{r}\overline{x}=\pm 3\frac{\sigma }{\sqrt{n}}\end{array}$

$=\pm 3\times 199.5$

$=\pm 598.5$

Asa result, the expected variation is $\pm 598.5$.