Year: 2023
Paper: 2
Question Number: 11
Course: LFM Stats And Pure
Section: Normal Distribution
No solution available for this problem.
Many candidates were able to express their reasoning clearly and presented good solutions to the questions that they attempted. There were excellent solutions seen for all of the questions. An area where candidates struggled in several questions was in the direction of the logic that was required in a solution. Some candidates failed to appreciate that separate arguments may be needed for the "if" and "only if" parts of a question and, in some cases, candidates produced correct arguments, but for the wrong direction. In several questions it was clear that candidates who used sketches or diagrams generally performed much better that those who did not. Sketches often also helped to make the solution clearer and easier to understand. Several questions on the STEP papers ask candidates to show a given result. Candidates should be aware that there is a need to present sufficient detail in their solutions so that it is clear that the reasoning is well understood.
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
\begin{questionparts}
\item $X_1$ and $X_2$ are both random variables which take values $x_1, x_2, \ldots, x_n$, with probabilities $a_1, a_2, \ldots, a_n$ and $b_1, b_2, \ldots, b_n$ respectively.
The value of random variable $Y$ is defined to be that of $X_1$ with probability $p$ and that of $X_2$ with probability $q = 1-p$.
If $X_1$ has mean $\mu_1$ and variance $\sigma_1^2$, and $X_2$ has mean $\mu_2$ and variance $\sigma_2^2$, find the mean of $Y$ and show that the variance of $Y$ is $p\sigma_1^2 + q\sigma_2^2 + pq(\mu_1 - \mu_2)^2$.
\item To find the value of random variable $B$, a fair coin is tossed and a fair six-sided die is rolled. If the coin shows heads, then $B = 1$ if the die shows a six and $B = 0$ otherwise; if the coin shows tails, then $B = 1$ if the die does \textbf{not} show a six and $B = 0$ if it does.
The value of $Z_1$ is the sum of $n$ independent values of $B$, where $n$ is large.
Show that $Z_1$ is a Binomial random variable with probability of success $\frac{1}{2}$.
Using a Normal approximation, show that the probability that $Z_1$ is within $10\%$ of its mean tends to $1$ as $n \longrightarrow \infty$.
\item To find the value of random variable $Z_2$, a fair coin is tossed and $n$ fair six-sided dice are rolled, where $n$ is large. If the coin shows heads, then the value of $Z_2$ is the number of dice showing a six; if the coin shows tails, then the value of $Z_2$ is the number of dice not showing a six.
Use part \textbf{(i)} to write down the mean and variance of $Z_2$.
Explain why a Normal distribution with this mean and variance will not be a good approximation to the distribution of $Z_2$.
Show that the probability that $Z_2$ is within $10\%$ of its mean tends to $0$ as $n \longrightarrow \infty$.
\end{questionparts}
There were very few substantial attempts at this question overall. In part (i) a large number of candidates incorrectly stated that Y = pX1 + qX2. However, there were several good responses to this question with many candidates obtaining the correct value for at least one of the mean and variance of Y. Similarly, in part (ii) many candidates were able to compute the mean and variance of Z1 correctly. However, several candidates only computed P(B = 1) when asked to justify that Z1 is a binomial variable. Candidates generally struggled with part (iii), often comparing the variance and the mean incorrectly for the two facts that were required to be shown.