Mathematics 9709/62 — May/June 2013

Walkthrough

We are told that the normal variable $Y$ has mean equal to five times its standard deviation, so $\mu = 5\sigma$. This means that once we know either $\mu$ or $\sigma$, we know both; the probability statement will let us find that single unknown.

The given probability $\mathrm{P}(Y > 20) = 0.0732$ describes a right tail. Since $0.0732$ is less than $0.5$, the boundary $20$ must be greater than the mean, and the corresponding standard normal $z$-value is positive. We find $z$ by looking up the value that leaves $0.0732$ in the upper tail, equivalently $0.9268$ in the lower tail. This gives $z = 1.452$.

Next, we write the standardisation formula:

Because $\mu = 5\sigma$, we replace $\sigma$ by $\frac{\mu}{5}$:

This is now a single linear equation in $\mu$. Multiplying through by $\frac{\mu}{5}$ gives $0.2904\mu = 20 - \mu$, so $1.2904\mu = 20$, and therefore $\mu = \frac{20}{1.2904} = 15.5$.

Key Takeaways

• The normal probability statement can be converted into a $z$-score using the inverse normal table.
• A right-tail probability less than $0.5$ corresponds to a positive $z$-score.
• Relationships between the mean and standard deviation reduce the number of unknowns and allow a direct equation to be solved.
• The standardisation formula $z = \frac{x - \mu}{\sigma}$ remains the key link between a normal variable and the standard normal distribution.

Common Mistakes

• Using $z = 0.0732$ directly instead of the inverse normal value $1.452$. The probability is not a $z$-score.
• Using the lower-tail probability $0.0732$ and getting a negative $z$-value, or forgetting that $\mathrm{P}(Y > 20)$ is a right tail.
• Treating the statement "mean is five times the standard deviation" as $\sigma = 5\mu$ rather than $\mu = 5\sigma$.
• Confusing variance $\sigma^2$ with standard deviation $\sigma$ in the standardisation formula.

Things to Be Careful About

• Check the direction of the tail. Since $0.0732 < 0.5$, the boundary is above the mean, so $z$ must be positive.
• Read the normal tables carefully: either find the upper-tail value directly or use $1 - 0.0732 = 0.9268$ in the body of the table.
• When rearranging $1.452 = \frac{20 - \mu}{\mu / 5}$, keep the algebra exact and avoid rounding until the final step.
• The mark scheme accepts $z = 1.45$ rounded, and the final mean should round to $15.5$; use sufficient accuracy in intermediate values.