Estimate the variance from the biggest and the smallest data

Question

What is the estimation that we can use? The answer is n=183.

Answer 1

Let $\bar{\sigma}^2$ the variance of the sample mean and $\sigma^2$ be the population mean. Then they are related by $\bar{\sigma}^2=\frac{\sigma^2}{N}$. So we can estimate N as $$N=\frac{\sigma^2}{\bar{\sigma}^2}$$.

Use the range estimate for $\sigma$, namely $$\sigma \approx \frac{\max-\min}{4}=\frac{12-5}{4}=\frac{7}{4}$$

Now since the estimate is within 10 seconds ($\frac{1}{6}$ minutes) with 99% confidence and $z_{0.005} = 2.576$, the solution takes $$2.576 \bar{\sigma}= \mathbf{2} \times \frac{1}{6}$$ Solving for $N$ from the estimates of $\sigma$ and $\bar{\sigma}$ above gives $N=182.8986$ which is your book answer.

Note: But I find this interpretation of "within 10 secs" to be strange. I think the correct way is to not have the factor of 2 that I have bolded in the last equation. The RHS should be just $\frac{1}{6}$, giving an estimate of 4 times as big, $N=732$