Mathematics 9709/53 — May/June 2025

Walkthrough

First, recall that the mean of a set of values is the total divided by the number of values. Since the mean of the 40 values is 124.0, the total $\sum x = 40 \times 124.0 = 4960$.

Next, look at the given sum $\sum (x - k)$. When you expand this sum, you get $\sum x - 40k$ because subtracting $k$ from each of the 40 values subtracts $40k$ in total. We are told this equals 836.0.

So we have the equation $4960 - 40k = 836$. Rearranging gives $40k = 4124$, so $k = 103.1$.

Key Takeaways

• The mean $\bar{x} = \frac{\sum x}{n}$.
• $\sum (x - k) = \sum x - nk$ is a useful identity when data is coded by subtracting a constant.
• The mean of the coded values is the original mean minus $k$.

Common Mistakes

• Forgetting to multiply the mean by 40 to get the total.
• Sign errors when rearranging $4960 - 40k = 836$ (e.g., writing $40k = 836 - 4960$).
• Not showing the intermediate equation; the mark scheme requires the equation $4960 - 40k = 836$ (or equivalent) for the first mark.

Things to Be Careful About

• The mark scheme accepts several equivalent forms: $124 - k = 20.9$ or $40k = 4124$, so any correct rearrangement is fine.
• The final answer should be given as a decimal, $k = 103.1$.

Walkthrough

The key idea is that adding or subtracting a constant from every data value does not change the spread of the data. So the variance (and standard deviation) of $x$ is the same as the variance of $(x - k)$.

We have the coded sums $\sum (x - k) = 836.0$ and $\sum (x - k)^2 = 25410.8$. The variance formula for a set of values $y$ is $\text{Var}(y) = \frac{1}{n}\sum y^2 - (\bar{y})^2$. Here $y = x - k$ and $n = 40$.

First find the mean of the coded values: $\frac{836.0}{40} = 20.9$.

Then:

The standard deviation is the square root of the variance:

An alternative method is to first expand $\sum (x - k)^2$ to find $\sum x^2 = 622978.4$, then use $\text{Var}(x) = \frac{622978.4}{40} - 124^2 = 198.46$. Both give the same result.

Key Takeaways

• Variance is invariant under adding or subtracting a constant: $\text{Var}(x) = \text{Var}(x - k)$.
• The variance formula $\text{Var} = \frac{1}{n}\sum y^2 - (\bar{y})^2$ works with coded data.
• Standard deviation is the positive square root of the variance.

Common Mistakes

• Using $\frac{\sum (x - k)}{40}$ instead of its square in the variance formula — you must subtract the square of the mean of the coded values.
• Forgetting to take the square root at the end.
• Confusing variance with standard deviation.

Things to Be Careful About

• The mark scheme requires the answer to at least 3 significant figures: 14.08758... so 14.1 is correct.
• If you use Method 2, make sure the expansion of $\sum (x - k)^2$ is correct: $\sum x^2 = 622978.4$.
• The standard deviation is positive; never give a negative value.

Walkthrough

We first identify from the stem that 45% of Science students study Drama, so $p = 0.45$. With 3 students selected at random from a large college, each selection is independent and has the same probability, so the number of Drama students follows a binomial distribution: $X \sim \mathrm{B}(3, 0.45)$.

The phrase "at least 1" means $X = 1, 2$ or $3$. Rather than adding three binomial probabilities, it is easier to use the complement: find the probability that none of the 3 students studies Drama, then subtract from 1. The probability a student does not study Drama is $1 - 0.45 = 0.55$, so $P(X = 0) = (0.55)^3$. Therefore $P(X \geq 1) = 1 - (0.55)^3$.

Key Takeaways

• For "at least one" events, the complement rule is usually the quickest method.
• A large population means the probability stays constant and selections are independent, so a binomial model can be used.
• Convert percentages to decimals before multiplying.

Common Mistakes

• Giving $0.45^3$ as the answer; that is the probability that all 3 study Drama, not at least 1.
• Forgetting to subtract from 1 after computing $P(X = 0)$.
• Treating "at least 1" as exactly 1.

Things to Be Careful About

• Use $0.55 = 1 - 0.45$ for the probability of no Drama.
• The mark scheme accepts the exact fraction $\frac{6669}{8000}$ or the rounded value 0.834. Do not round intermediate values too early.

Walkthrough

From the stem, the probability that a Science student studies Art is 0.20 and the probability that one studies Music is 0.35. These categories are mutually exclusive, so the probability of studying Art or Music is $0.20 + 0.35 = 0.55$.

Let $Y$ be the number of students who study Art or Music in a random sample of 10. Because the college is large, $Y$ follows a binomial distribution: $Y \sim \mathrm{B}(10, 0.55)$.

The event "more than 7" includes $Y = 8$, $Y = 9$ and $Y = 10$. We use the binomial formula

for $r = 8, 9, 10$, and add the three results because the outcomes are mutually exclusive.

Key Takeaways

• Mutually exclusive categories can be combined by adding their probabilities.
• "More than 7" means 8, 9 and 10, not 7.
• The binomial formula is applied to each favourable value of $r$ and the results are summed.

Common Mistakes

• Using $p = 0.45$ (Drama) instead of $p = 0.55$ (Art or Music).
• Including $Y = 7$ in the sum; "more than 7" means $Y = 8, 9, 10$.
• Forgetting the binomial coefficient ${}^{10}C_r$.
• Rounding each term too early, which can move the final answer outside the accepted range.

Things to Be Careful About

• For $r = 10$, the binomial coefficient is 1, so the term is simply $(0.55)^{10}$.
• The accepted final answer is 0.0996, or any value in the range $0.09955 < p \leq 0.0996$.
• Keep enough decimal places in intermediate terms to avoid losing accuracy.

Walkthrough

$X$ is the number of throws until a 3 first appears. Since the die is fair, $p = \frac{1}{6}$ and $q = \frac{5}{6}$. For $X = 8$, the first seven throws must all be non-3s and the eighth throw must be a 3. The probability of seven independent failures is $q^7$, and we multiply by $p$ for the final success. This gives the geometric formula $P(X = k) = q^{k-1}p$.

Key Takeaways

• The number of trials until the first success follows a geometric distribution.
• For a geometric distribution, $P(X = k) = q^{k-1}p$.
• The success must occur on the last of the $k$ throws, so the exponent on $q$ is $k-1$.

Common Mistakes

• Writing $\left(\frac{1}{6}\right)^8$ instead of $\left(\frac{1}{6}\right)\left(\frac{5}{6}\right)^7$.
• Using the wrong exponent on $\frac{5}{6}$.
• Not showing the calculation, which can lose the method mark.

Things to Be Careful About

• $X = 8$ means seven failures followed by one success.
• The answer may be given as a fraction or a decimal; $0.0465$ is accepted.
• Each throw is independent, so probabilities multiply.

Walkthrough

$X < 9$ means the first success occurs on throw 1, 2, ..., or 8. The direct method would sum eight geometric probabilities. The quicker method is to notice that the event $X < 9$ fails only if no 3 appears in the first 8 throws. The probability of no 3 in 8 independent throws is $\left(\frac{5}{6}\right)^8$. Therefore $P(X < 9) = 1 - \left(\frac{5}{6}\right)^8$.

Key Takeaways

• Complements often simplify geometric sums.
• $P(X < 9) = P(X \leq 8) = 1 - q^8$.
• Check whether the inequality is strict or inclusive.

Common Mistakes

• Writing $P(X < 9) = \left(\frac{5}{6}\right)^8$; that gives the probability of no success in 8 throws, not the required probability.
• Using the exponent 9 instead of 8.
• Omitting the $1 -$ step and losing the method mark.

Things to Be Careful About

• $X < 9$ includes $X = 1$ through $X = 8$, not $X = 9$.
• The answer should be 0.767 (AWRT).
• Show the complement method clearly to earn the M1 mark.

Walkthrough

We need the second 3 to occur before the 6th throw, so it must occur on throw 2, 3, 4 or 5. For each possible throw $n$, the second 3 is on throw $n$, so exactly one 3 appears among the first $n-1$ throws and the $n$th throw is a 3. The earlier 3 can be placed in $\binom{n-1}{1}$ positions. Thus the probability for a given $n$ is $\binom{n-1}{1}\left(\frac{1}{6}\right)^2\left(\frac{5}{6}\right)^{n-2}$. These four events are mutually exclusive, so adding them gives the required probability.

An alternative is to use the complement: the second 3 is not before the 6th throw if, in the first 5 throws, there are 0 or 1 threes. So

Key Takeaways

• When a fixed number of successes must occur by a certain number, split the cases by the trial on which the last required success occurs.
• Use a binomial coefficient to count which of the earlier trials contains the previous success.
• Mutually exclusive events can be summed.

Common Mistakes

• Interpreting 'before the 6th throw' as the second 3 occurring on throw 6.
• Forgetting the binomial coefficient that places the first 3.
• Using $\left(\frac{5}{6}\right)^{n-1}$ instead of $\left(\frac{5}{6}\right)^{n-2}$, since there are exactly two successes and $n-2$ failures in $n$ throws.
• Not adding all four cases.

Things to Be Careful About

• The second 3 cannot occur on throw 1, so the smallest possible case is $n = 2$.
• In the complement method, subtract both 'no 3 in 5 throws' and 'exactly one 3 in 5 throws'.
• The mark scheme accepts answers in the range $0.196 \leq p < 0.1963$; the exact fraction is $\frac{763}{3888}$.
• If using the complement method, remember the factor $\binom{5}{1}=5$ for the position of the single 3.

Walkthrough

The question asks for the number of people who completed the puzzle between 4 and 7.5 minutes. On a cumulative frequency graph, the value on the y-axis at a given x-value represents the total number of people who completed the puzzle in that time or less. To find the number of people in a specific interval $[t_1, t_2]$, we read the cumulative frequency at $t_2$ and subtract the cumulative frequency at $t_1$. Reading from the graph, at $t=4$, the cumulative frequency is about 25. At $t=7.5$, it is about 68 or 69. Subtracting 25 from these values gives 43 or 44.

Key Takeaways

Cumulative frequency graphs allow us to estimate the number of observations within any time interval by finding the difference in cumulative frequencies at the interval endpoints.

Common Mistakes

• Reading the wrong axis values.
• Forgetting to subtract the lower cumulative frequency from the higher one.
• Accepting answers outside the reasonable range (43 or 44) due to poor graph reading.

Things to Be Careful About

Always ensure you are reading the cumulative frequency (y-axis) for the given time values (x-axis). Small variations in reading the graph are expected, so answers like 43 or 44 are both acceptable.

Walkthrough

A box-and-whisker plot summarises data using five key values: minimum, lower quartile (LQ), median, upper quartile (UQ), and maximum. These can be read from a cumulative frequency graph.

1. Minimum and Maximum: The graph starts at $t=2$ (CF=0) and ends at $t=12$ (CF=84), so the minimum is 2 and maximum is 12.
2. Median: Half of 84 is 42. Find 42 on the y-axis, move horizontally to the curve, then down to the x-axis to get $t \approx 4.8$.
3. Lower Quartile: One quarter of 84 is 21. Find 21 on the y-axis, move to the curve, then down to get $t \approx 3.7$ or $3.8$.
4. Upper Quartile: Three quarters of 84 is 63. Find 63 on the y-axis, move to the curve, then down to get $t \approx 6.5$ or $6.6$.

Finally, draw these on the grid with a linear scale, ensuring the axis is labelled and whiskers are drawn correctly (not through the box).

Key Takeaways

Box-and-whisker plots provide a visual summary of the spread and central tendency of data. The five-number summary is directly readable from cumulative frequency graphs.

Common Mistakes

• Using a non-linear scale for the box-and-whisker plot.
• Drawing whiskers through the box or at the corners.
• Forgetting to label the axis with units ("Time in minutes").
• Incorrect quartile positions (e.g., using $N/4$ and $3N/4$ incorrectly).

Things to Be Careful About

Ensure the scale is linear and covers at least the range from 2 to 12. The whiskers should extend from the minimum to the LQ and from the UQ to the maximum, without passing through the box. Acceptable variations in reading the graph (e.g., LQ=3.7 or 3.8) are allowed.

Walkthrough

We model this two-step process using a probability tree diagram. The first node splits into three branches representing the marble drawn from Bag A: red (R), blue (B), or green (G). Since Bag A has 12 marbles total (6 red, 5 blue, 1 green), the branch probabilities are $\frac{6}{12}$, $\frac{5}{12}$, and $\frac{1}{12}$.

After the first marble is transferred to Bag B, the second draw is from Bag B, which now contains 9 marbles. The composition of Bag B depends on what was added:

• Adding a red marble gives Bag B: 6 red, 3 blue (9 total). The probabilities for the second draw are $\frac{6}{9}$ for red and $\frac{3}{9}$ for blue.
• Adding a blue marble gives Bag B: 5 red, 4 blue (9 total). The probabilities are $\frac{5}{9}$ for red and $\frac{4}{9}$ for blue.
• Adding a green marble gives Bag B: 5 red, 3 blue, 1 green (9 total). The probabilities are $\frac{5}{9}$ for red, $\frac{3}{9}$ for blue, and $\frac{1}{9}$ for green.

Each branch of the tree is labelled with the corresponding probability. The tree diagram captures all possible outcomes and their probabilities.

Key Takeaways

• A probability tree diagram is a visual tool for representing sequential random events.
• At each stage, the probabilities on the branches must sum to 1.
• Conditional probabilities at the second stage depend on the outcome of the first stage, because the composition of the second bag changes.

Common Mistakes

• Forgetting to update the contents of Bag B before calculating second-stage probabilities.
• Writing the second-stage probabilities as if Bag B still has 8 marbles instead of 9.
• Omitting the green branch from Bag B when a green marble is transferred (probability $\frac{1}{9}$).

Things to Be Careful About

• The first set of branches must always represent Bag A, as that is the first event in the sequence.
• Branches with zero probability (e.g., drawing green from Bag B when no green was added) may be omitted or shown with probability 0, but additional branches beyond R, B, G lose the mark.
• Always verify that probabilities on branches from each node sum to 1.

Walkthrough

We want the probability that both marbles chosen are the same colour. Looking at the tree diagram, there are exactly three paths that give matching colours:

1. Both red (RR): Draw red from Bag A ($\frac{6}{12}$), then draw red from Bag B ($\frac{6}{9}$). Probability = $\frac{6}{12} \times \frac{6}{9} = \frac{36}{108}$.

2. Both blue (BB): Draw blue from Bag A ($\frac{5}{12}$), then draw blue from Bag B ($\frac{4}{9}$). Probability = $\frac{5}{12} \times \frac{4}{9} = \frac{20}{108}$.

3. Both green (GG): Draw green from Bag A ($\frac{1}{12}$), then draw green from Bag B ($\frac{1}{9}$). Probability = $\frac{1}{12} \times \frac{1}{9} = \frac{1}{108}$.

These three events cannot happen simultaneously, so they are mutually exclusive. We add their probabilities:

Key Takeaways

• The multiplication law gives the probability of a specific sequence of events (following a path on the tree).
• The addition law is used to combine probabilities of mutually exclusive outcomes.
• Always check which paths on the tree correspond to the desired event.

Common Mistakes

• Forgetting the GG path (it is easy to overlook since the green marble is rare).
• Using the wrong denominator for the second stage (should be 9, not 8).
• Not simplifying the final fraction.

Things to Be Careful About

• Ensure all three same-colour combinations (RR, BB, GG) are included.
• The working must show the full expression $\frac{6}{12} \times \frac{6}{9} + \frac{5}{12} \times \frac{4}{9} + \frac{1}{12} \times \frac{1}{9}$ to earn the method mark.

Walkthrough

We are asked for the probability that the marble from Bag A is blue, given that the marble from Bag B is blue. This is a conditional probability problem.

The conditional probability formula is:

Here, $A$ is the event 'marble from Bag A is blue' and $B$ is the event 'marble from Bag B is blue'.

Numerator $P(A \cap B)$: This is the probability that both marbles are blue, i.e., path BB on the tree:

Denominator $P(B)$: We need the total probability that the marble from Bag B is blue, regardless of what was drawn from Bag A. There are three paths that lead to blue from Bag B:

1. Red from A, then blue from B: $\frac{6}{12} \times \frac{3}{9} = \frac{18}{108}$
2. Blue from A, then blue from B: $\frac{5}{12} \times \frac{4}{9} = \frac{20}{108}$
3. Green from A, then blue from B: $\frac{1}{12} \times \frac{3}{9} = \frac{3}{108}$

Summing these:

Final calculation:

Key Takeaways

• Conditional probability requires identifying the intersection event (numerator) and the conditioning event (denominator).
• The denominator is often found using the law of total probability by summing over all paths that lead to the conditioning event.
• Fractions with the same denominator simplify cleanly.

Common Mistakes

• Using only one path in the denominator instead of summing all paths where Bag B gives blue.
• Forgetting that the green-from-A path also contributes to blue-from-B ($\frac{1}{12} \times \frac{3}{9}$).
• Dividing incorrectly or not cancelling the common denominator.

Things to Be Careful About

• The mark scheme awards a method mark for showing the numerator as $\frac{5}{12} \times \frac{4}{9}$ (or $\frac{20}{108}$) and a separate mark for the correct denominator expression. Both must be shown clearly.
• Acceptable decimal answers are 0.488 or 0.4878, but the exact fraction $\frac{20}{41}$ is preferred.

Walkthrough

We are told that large bags of pasta have masses that are normally distributed with mean $2.50$ kg and standard deviation $0.12$ kg. To find the probability that a randomly chosen bag has mass less than $2.65$ kg, we convert $2.65$ to a standard normal $Z$-score. The $Z$-score tells us how many standard deviations $2.65$ is above the mean. Here $2.65 - 2.50 = 0.15$, and dividing by $0.12$ gives $1.25$. We then look up $\text{P}(Z < 1.25)$ in the standard normal table, obtaining $0.8944$. This is the required probability.

Key Takeaways

The key idea is that any normal distribution can be converted to the standard normal distribution using $Z = \frac{X - \mu}{\sigma}$. Once converted, probabilities are read from the standard normal table. Always subtract the mean before dividing by the standard deviation.

Common Mistakes

• Forgetting to subtract the mean before dividing by the standard deviation.
• Using the wrong tail of the distribution. Here we need the probability less than $2.65$, so we use the cumulative probability from the left.
• Reading the standard normal table incorrectly for $z = 1.25$.

Things to Be Careful About

The standard deviation is $0.12$ kg, not the variance. The mark scheme allows some flexibility with $\sigma^2$ or $\sqrt{\sigma}$ in part (a), but it is safest to use the given standard deviation directly. No continuity correction is needed because the variable is continuous.

Walkthrough

We first translate “more than $1.65$ standard deviations above the mean” into a concrete mass: $2.50 + 1.65 \times 0.12 = 2.698$ kg. Standardising this mass gives $Z = 1.65$, so the required probability is $\text{P}(Z > 1.65)$. The standard normal table gives the left-tail probability $\text{P}(Z < 1.65) = 0.9505$, so the right-tail probability is $1 - 0.9505 = 0.0495$. This is the probability that one randomly chosen large bag is more than $1.65$ standard deviations above the mean. For $160$ bags, the expected number is $160 \times 0.0495 = 7.92$. Since the question asks for a number of bags, we give a whole-number answer; the exact expected value is $7.92$, so the integer answer is $8$ (the mark scheme also allows $7$ or $8$).

Key Takeaways

• “$1.65$ standard deviations above the mean” means the mass $\mu + 1.65\sigma$.
• For a sample of $n$ items, the expected number with a given property is $n \times p$, where $p$ is the probability for one item.

Common Mistakes

• Interpreting $1.65$ as a mass instead of a $Z$-score, or adding $1.65$ to the mean without multiplying by the standard deviation.
• Forgetting to subtract the left-tail probability from $1$ when finding $\text{P}(Z > 1.65)$.
• Giving the final answer as $7.92$ when the question asks for a number of bags; the mark scheme requires a whole-number answer.

Things to Be Careful

• Use the standard deviation $0.12$, not $\sigma^2$ or $\sqrt{\sigma}$.
• No continuity correction is used because this is a normal distribution calculation.
• When following through from a calculated probability, use at least 4 decimal places before multiplying by $160$.

Walkthrough

We need to find $\mu$ and $\sigma$ from two percentage conditions. First, $77\%$ of bags have mass greater than $1.26$ kg, so $23\%$ have mass less than $1.26$ kg. The standard normal table tells us that a left-tail probability of $0.23$ corresponds to $z = -0.739$. Thus $\frac{1.26 - \mu}{\sigma} = -0.739$. Similarly, $44\%$ of bags have mass less than $1.35$ kg, and the corresponding $z$-value is $-0.151$, giving $\frac{1.35 - \mu}{\sigma} = -0.151$. These are two linear equations in $\mu$ and $\sigma$. Rearrange each to $\mu = 1.26 + 0.739\sigma$ and $\mu = 1.35 + 0.151\sigma$, equate them, and solve for $\sigma$. Then substitute back to find $\mu$. The answers are $\mu = 1.37$ kg and $\sigma = 0.153$ kg.

Key Takeaways

• Percentages in normal distribution questions must be converted to probabilities before looking up $z$-values.
• The inverse normal table gives a $z$-value for a given cumulative probability.
• Two conditions on a normal distribution produce two equations in $\mu$ and $\sigma$, which can be solved simultaneously.

Common Mistakes

• Using the percentages $0.77$, $0.23$, $0.44$ directly as $z$-values instead of looking up the inverse normal values.
• Mixing up the tail direction. For $77\%$ greater than $1.26$, we use $23\%$ less than $1.26$.
• Using the wrong sign for the $z$-values; both $z$-values here are negative.
• Making an algebraic error when rearranging the standardisation equations.

Things to Be Careful

• Use $\sigma$, not $\sigma^2$ or $\sqrt{\sigma}$, in the standardisation formula.
• The mark scheme allows $z_1$ between $-0.740$ and $-0.738$, and $z_2$ between $-0.152$ and $-0.150$.
• The final values are $\mu = 1.37$ and $\sigma = 0.153$ (AWRT).

Walkthrough

The phrase "not stand next to each other" suggests counting the complement. First count all arrangements of 7 distinct men: $7!$. Then count the arrangements where Ali and Ben are together. To do this, glue Ali and Ben into one block, so there are 6 objects (the block plus the other 5 men), giving $6!$ arrangements. Inside the block Ali and Ben can be in either order, so multiply by $2$. Subtract this from the total to leave exactly the arrangements where they are not adjacent.

Key Takeaways

This question uses the "total minus unwanted" technique for restrictions. When two people must be together, treat them as a single block and multiply by the internal arrangements.

Common Mistakes

Forgetting to multiply by $2$ for the order of Ali and Ben inside the block. Alternatively, forgetting to subtract from the total and just giving $6! \times 2$.

Things to Be Careful About

The block method counts arrangements where Ali and Ben are together. Since Ali and Ben are distinct, their internal order matters. The final answer must be an integer; no double-counting occurs because the cases "together" and "not together" are complementary.

Walkthrough

The condition "all the men stand together and all the women stand together" means we can think of the men as one block and the women as one block. First arrange the two blocks: the men's block could come before the women's block, or the women's block before the men's block, giving $2!$ arrangements. Then arrange the 7 men inside their block in $7!$ ways and the 4 women inside theirs in $4!$ ways. Multiplying gives the total.

Key Takeaways

Blocking is the standard technique when a group of items must stay together. The order of the blocks and the internal order within each block must both be counted.

Common Mistakes

Forgetting to arrange the two blocks, or forgetting to arrange the people inside one of the blocks.

Things to Be Careful About

The blocks are distinct because one contains men and one contains women, so there is no division by $2!$. The answer is large: $241920$.

Walkthrough

The groups have different sizes, so they are distinct: a group of 6, a group of 3 and a group of 2. Choose the 6 people for the first group from the 11: $\binom{11}{6}$. From the remaining 5 choose the 3 for the second group: $\binom{5}{3}$. The last 2 are forced. Multiplying gives the number of ways.

Key Takeaways

When dividing into groups of different sizes, choose groups sequentially; the groups are distinguished by their sizes, so no division by a factorial is needed.

Common Mistakes

Dividing by $3!$ because the groups are labelled by size, or using permutations instead of combinations when order within a group does not matter.

Things to Be Careful About

The order of choosing the groups does not affect the count as long as the sizes are fixed. The final answer is $4620$.

Walkthrough

We need the probability that Ali, Ben and Charlie are in the same group. The group sizes are 6, 3 and 2, so the three brothers cannot all be in the group of 2. They must therefore be together either in the group of 6 or in the group of 3.

For the group of 6: place Ali, Ben and Charlie in it, then choose the other 3 members from the remaining 8 people: $\binom{8}{3}=56$. The remaining 5 people are then split into a group of 3 and a group of 2: $\binom{5}{3}\binom{2}{2}=10$. So this case gives $560$ favourable groupings.