The time, in minutes, taken by men to run a marathon is modelled by a normal distribution with mean 240 minutes and standard deviation 40 minutes - Edexcel - A-Level Maths Statistics - Question 6 - 2016 - Paper 1

To estimate Nathaniel's longest time to finish within the first 20% of male runners, we need to find the corresponding z-score for the 20th percentile:

From z-tables, we find that the z-score for the 20th percentile is approximately $-0.8416$.

Using the z-score formula:

$Z = \frac{X - \mu}{\sigma}$

Now, we rearrange it to find $X$:

$X = Z \cdot \sigma + \mu$

Substituting the values:

$X = (-0.8416) \cdot 40 + 240$ $X \approx -33.664 + 240 = 206.336$

Therefore, Nathaniel should aim to finish in approximately 206 minutes.

To find the probability $P(W < \mu - 30 | W < \mu)$ using the provided information, we know:

$P(W < \mu + 30) = 0.82$

This implies:

$P(W < \mu) = P(W < \mu + 30) - P(W \in (\mu, \mu + 30)) = 0.82 - P(W > \mu)$

Let’s denote: $P(W < \mu) = P(W < \mu) = q$ $P(W < \mu - 30) = P(W < \mu - 30 | W < \mu) \cdot P(W < \mu)$

Using the z-score for $\mu - 30$ calculations, we have:

• $\phi(-1) = P(W < \mu - 30)$
• and $\sigma$ for standard deviation leads to the conclusion for interpretation.

Thus, computing provides: $P(W < \mu - 30 | W < \mu) = \frac{0.36}{0.82} \approx 0.36$

So, $P(W < \mu - 30 | W < \mu)$ is approximately 0.36.