Mathematics 9709/53 — October/November 2020

Walkthrough

We are told that the swimming time has a normal distribution with mean $62$ s and standard deviation $5$ s. The normal curve is symmetric about the mean, and the total area under it is $1$. We want the area between $56$ and $66$ seconds, which is the probability that a random member's time lies in that interval.

First convert the raw times to $z$-scores using $z = \frac{X - \mu}{\sigma}$, with $\mu = 62$ and $\sigma = 5$. For $X = 56$, $z = -1.2$; for $X = 66$, $z = 0.8$. So the required probability is $P(-1.2 < Z < 0.8)$.

Next use the standard normal table. Directly, $P(-1.2 < Z < 0.8) = \Phi(0.8) - \Phi(-1.2)$. Since tables usually give only positive $z$-values, use $\Phi(-1.2) = 1 - \Phi(1.2)$, which gives $\Phi(0.8) + \Phi(1.2) - 1$. Substituting $0.7881 + 0.8849 - 1 = 0.673$. The required probability is therefore $0.673$.

Key Takeaways

This question tests standardisation of a normal variable and the use of the standard normal table. The key skill is to convert raw limits into $z$-scores, then identify the correct area using symmetry when negative $z$-values are involved.

Common Mistakes

• Forgetting to standardise the bounds, or standardising only one bound.
• Using $\sigma^2$ or $\sqrt{\sigma}$ in the denominator instead of the standard deviation $5$.
• Using the wrong tail, e.g. computing only the area above $66$ or below $56$.
• Forgetting that $\Phi(-1.2) = 1 - \Phi(1.2)$ when the table does not give negative $z$-values.

Things to Be Careful About

• Use the standard deviation $5$ in the denominator, not the variance $25$.
• The distribution is already continuous, so a continuity correction is not needed.
• Read the table accurately: $\Phi(0.8) = 0.7881$ and $\Phi(1.2) = 0.8849$.

Walkthrough

We need the time $t$ minutes such that 13% of members take longer than that. Since the cutoff is above the mean, the corresponding $z$-score should be positive.

The upper-tail probability is $0.13$, so the lower-tail probability up to the cutoff is $1 - 0.13 = 0.87$. From the standard normal table, the $z$-value with lower-tail area $0.87$ is $z = 1.127$ (within the allowed tolerance $1.126$ to $1.127$).

Now convert the cutoff $t$ minutes into seconds: the cutoff time is $60t$ seconds. Standardise using the mean $62$ and standard deviation $5$:

$$\frac{60t - 62}{5} = 1.127$$

Multiply through by 5:

$$60t - 62 = 5.635$$

Add 62:

$$60t = 67.635$$

Divide by 60:

$$t = 1.12725$$

which rounds to $t = 1.13$ minutes.

Key Takeaways

This part tests the inverse use of the normal distribution: given a probability, find the corresponding value. It also checks unit conversion and algebraic solving of the z-score formula.

Common Mistakes

• Forgetting to convert minutes to seconds, using $t - 62$ instead of $60t - 62$.
• Treating the 13% as a lower tail and using $z = -1.127$.
• Taking the z-value for 13% lower tail instead of upper tail.
• Not rounding the final answer correctly; the mark scheme requires $t = 1.13$ (CAO).

Things to Be Careful About

• The mean and standard deviation are in seconds; the final answer is required in minutes.
• Use the positive $z$-score $1.127$, because the time is above the mean.
• Ensure the final answer is to three significant figures: $1.13$ minutes.