Mathematics 9709/51 — May/June 2024

Walkthrough

The question provides summary statistics for two separate groups of data, each coded by subtracting 30 from every value. The first group has 20 values with $\sum(x-30) = 439$, and the second group has 25 values with $\sum(x-30) = 470$.

To find the mean of all 45 values combined, we first need the total sum of $(x-30)$ across both groups. This is simply the sum of the two given sums: $439 + 470 = 909$.

Next, we divide this total by the total number of values, which is $20 + 25 = 45$. This gives the mean of the coded values: $909/45 = 20.2$.

Since every value was reduced by 30 in the coding, the mean of the original values $x$ is the mean of $(x-30)$ plus 30: $20.2 + 30 = 50.2$.

Key Takeaways

• The mean of a coded variable $(x - c)$ is the mean of $x$ minus $c$.
• When combining two groups, add the total sums before dividing by the total count.
• The total number of values is $20 + 25 = 45$.

Common Mistakes

• Forgetting to add 30 back to find the mean of $x$.
• Dividing each group by its own count instead of the combined total.
• Using 40 instead of 45 as the total count.

Things to Be Careful About

The total count is $20 + 25 = 45$, not 40. Always add the coding sums before dividing. The mean of $(x-30)$ is 20.2, so the mean of $x$ is 50.2.

Walkthrough

The standard deviation is the square root of the variance. The variance of $x$ equals the variance of $(x-30)$ because subtracting a constant from every value shifts the data but does not change its spread. We use the formula $\text{Var}(X) = E(X^2) - (E(X))^2$.

First, find the total sum of $(x-30)^2$ across all 45 values: $12405 + 11346 = 23751$. The mean of $(x-30)^2$ is $23751/45 = 527.8$. The mean of $(x-30)$ is $20.2$ (from part (a)), so its square is $20.2^2 = 408.04$.

The variance is then $527.8 - 408.04 = 119.76$. The standard deviation is the square root: $\sqrt{119.76} \approx 10.94 \approx 10.9$.

Key Takeaways

• Variance formula: $\text{Var}(X) = E(X^2) - (E(X))^2$.
• Variance is invariant under a coding shift: $\text{Var}(x) = \text{Var}(x-30)$.
• Standard deviation is the square root of the variance.

Common Mistakes

• Forgetting to subtract the square of the mean.
• Using the mean of $x$ (50.2) instead of the mean of $(x-30)$ (20.2) in the square.
• Not taking the square root at the end.

Things to Be Careful About

The variance is computed on the coded data, but the result is the same as for the original data. Use the mean of $(x-30)$, not the mean of $x$, in the formula. Round the final answer to 1 decimal place.

Walkthrough

We are told that tail lengths follow a normal distribution with mean 28 cm and standard deviation 3.3 cm. To find the probability that a randomly chosen tail length lies between 23 cm and 35 cm, we convert both raw values to $Z$-scores. The $Z$-score measures how many standard deviations a value lies from the mean.

For the lower bound 23 cm:

This means 23 cm is about 1.515 standard deviations below the mean. For the upper bound 35 cm:

This means 35 cm is about 2.121 standard deviations above the mean.

We need $P(-1.515 < Z < 2.121)$. Using the cumulative distribution function $\Phi$, this equals $\Phi(2.121) - \Phi(-1.515)$. Because the normal curve is symmetric about zero, $\Phi(-1.515) = 1 - \Phi(1.515)$, so the probability becomes $\Phi(2.121) + \Phi(1.515) - 1$. Looking up the table values $\Phi(2.121) \approx 0.9830$ and $\Phi(1.515) \approx 0.9351$ gives $0.9830 + 0.9351 - 1 = 0.9181 \approx 0.918$.

Key Takeaways

• Standardising converts any normal distribution to the standard normal distribution.
• The symmetry of the normal curve lets us convert negative $Z$-scores to positive ones.
• The probability between two values is the difference of the two cumulative probabilities.

Common Mistakes

• Forgetting to subtract the lower cumulative probability $\Phi(-1.515)$.
• Using $\Phi(-1.515)$ directly as a negative contribution instead of converting via symmetry.
• Reading the wrong row or column in the standard normal table.

Things to Be Careful About

• The mark scheme requires the standardisation formula to be shown explicitly for at least one bound.
• No continuity correction is applied here because this is a direct normal problem, not a binomial approximation.
• The final answer should be rounded to three significant figures (AWRT 0.918).

Walkthrough

We are told that 75% of raccoons have mass greater than 7.6 kg. Since the mean is 8.5 kg, the value 7.6 kg lies below the mean, so the corresponding $Z$-score must be negative.

We need the $Z$-score such that $P(Z > z) = 0.75$. From the standard normal table, $P(Z > 0.674) = 0.25$, so by symmetry $P(Z > -0.674) = 0.75$. Hence the critical value is $z = -0.674$.

Standardising the value 7.6 kg:

Setting this equal to $-0.674$:

Solving:

Key Takeaways

• Working backwards from a probability to an unknown parameter requires finding the critical $Z$-score first.
• The symmetry of the normal distribution is essential for handling probabilities greater than 0.5.
• The standardisation formula works in both directions: from value to $Z$-score, and from $Z$-score to value.

Common Mistakes

• Using $z = +0.674$ instead of $-0.674$, which would produce a negative value for $\sigma$.
• Confusing $P(Z > z)$ with $P(Z < z)$, leading to the wrong critical value.
• Using the probability 0.75 directly in the standardisation formula instead of the $Z$-score.

Things to Be Careful About

• The mark scheme requires the standardisation formula to be shown.
• The final answer should lie in the range $1.33 \leq \sigma \leq 1.34$.
• Recognise that a probability greater than 0.5 corresponds to a negative $Z$-score when the value is below the mean.

Walkthrough

First, calculate the frequency density for each class interval. Since the class widths are not uniform, we must use frequency density (frequency divided by class width) as the height of the bars to ensure the area of each bar is proportional to the frequency.

The class widths are: 20, 10, 10, 5, and 20. Dividing the frequencies by these widths gives frequency densities of 0.8, 3.2, 7.6, 12.8, and 0.6 respectively.

Next, draw the histogram on the provided grid. The horizontal axis represents height $h$ in cm, and the vertical axis represents frequency density. The bars must be drawn with the calculated heights and correct boundaries, with no gaps between adjacent bars.

Key Takeaways

• When class widths are unequal, frequency density must be used as the vertical axis in a histogram.
• Frequency density is calculated as $\frac{\text{frequency}}{\text{class width}}$.
• The area of each bar represents the frequency, so bar heights must be proportional to frequency density.

Common Mistakes

• Using frequency directly as the bar height instead of frequency density, which would make the area of the bars incorrect.
• Forgetting to label the vertical axis as frequency density or using an inappropriate scale.
• Drawing gaps between bars, which is incorrect for continuous data histograms.

Things to Be Careful About

• Ensure the horizontal scale covers the full range from 130 to 195 cm.
• The vertical scale must be large enough to accommodate the highest frequency density (12.8), so a scale up to at least 14 is appropriate.
• Axes must be clearly labelled with units where appropriate.

Walkthrough

To find the interquartile range, we first locate the class intervals containing the lower quartile (LQ) and upper quartile (UQ).

With a total of 200 data points, the LQ is the 50th value and the UQ is the 150th value.

Calculating cumulative frequencies:

• $130 \le h < 150$: 16
• $150 \le h < 160$: $16 + 32 = 48$
• $160 \le h < 170$: $48 + 76 = 124$
• $170 \le h < 175$: $124 + 64 = 188$
• $175 \le h < 195$: $188 + 12 = 200$

The 50th value is in the class $160 \le h < 170$, so the LQ $\ge 160$.
The 150th value is in the class $170 \le h < 175$, so the UQ $\le 175$.

The interquartile range is $R = \text{UQ} - \text{LQ}$. To find the maximum possible value of $R$, we use the largest possible UQ and the smallest possible LQ:

Thus, $R$ is not greater than 15.

Key Takeaways

• The interquartile range can be bounded using the class limits of the quartile classes even without exact quartile values.
• Cumulative frequencies help identify which class interval contains a given percentile.

Common Mistakes

• Calculating exact quartile values using interpolation when only an inequality is required.
• Misidentifying the class intervals for the LQ and UQ.

Things to Be Careful About

• The question asks to show $R$ is not greater than 15, so finding the maximum possible value is sufficient.
• Ensure cumulative frequencies are calculated correctly to locate the quartile positions.

Walkthrough

We think of the turn as up to three throws. The event of scoring no point in the turn is that all three throws are not 4. Since each throw is independent, the probability of this is $\left(\frac{3}{4}\right)^3$. The complement of this event is the required probability of obtaining a 4 in the turn.

Key Takeaways

• Use the complement rule: $P(\text{at least one success}) = 1 - P(\text{no success})$.
• For repeated independent throws, multiply the probability for each throw.

Common Mistakes

• Trying to add the probabilities for the first, second and third throws without dealing with the stopping rule.
• Forgetting that a turn still has up to three throws even if it ends early.

Things to Be Careful About

• The die is fair, so $P(4) = \frac{1}{4}$ and $P(\text{not }4) = \frac{3}{4}$.
• The maximum of three throws means the probability of no point is $\left(\frac{3}{4}\right)^3$.
• The answer must be shown as $\frac{37}{64}$.

Walkthrough

"First two turns" means two turns for Xeno and two turns for Yao, so four turns in total. From part (a), the probability of scoring no point in one turn is $\frac{27}{64}$. Because all four turns are independent, multiply this probability four times.

An equivalent view is that each no-point turn requires three throws that are not 4, so the probability is $\left(\frac{3}{4}\right)^{12}$.

Key Takeaways

• Use a previous result when it gives a useful probability.
• Multiply probabilities for independent events.

Common Mistakes

• Using only two turns instead of four.
• Using the success probability $\frac{37}{64}$ instead of the no-point probability $\frac{27}{64}$.

Things to Be Careful About

• The mark scheme allows follow-through from part (a): $\left(1 - \text{their}(a)\right)^4$.
• Accept unsimplified answers; the final value is approximately $0.0317$.

Walkthrough

Each player's score is the number of successful turns out of three, so it follows a binomial distribution with $n = 3$ and $p = \frac{37}{64}$. For Xeno to score 2 more points than Yao, the possible score pairs are $(3,1)$ and $(2,0)$. For each pair, multiply the probability of Xeno's score by the probability of Yao's score, since the two players are independent. The binomial coefficient counts the number of ways each score can occur. Then add the two mutually exclusive cases.

Key Takeaways

• Recognise a binomial distribution in repeated independent trials.
• Use binomial coefficients to count arrangements.
• Add probabilities of mutually exclusive cases.

Common Mistakes

• Forgetting the binomial coefficient $3$ in one of the cases.
• Considering only one of the two possible score pairs.
• Swapping $p$ and $q$.
• Multiplying the two cases together instead of adding them.

Things to Be Careful About

• The final answer is $0.0914$ rounded to 3 significant figures.
• The mark scheme awards a mark for one correct scenario, a method mark for adding two correct scenarios, and the final mark for the sum.
• Unsimplified expressions are accepted.

Walkthrough

Each day is an independent trial with two outcomes (snow or no snow) and the same probability of snow $p = 0.7$. This makes the number of snowy days in a fixed number of days a binomial random variable. Here $n = 7$ (a week) and $p = 0.7$.

We want "at least 5 days", which means $X = 5$, $X = 6$, or $X = 7$. Because these outcomes are mutually exclusive, we add their individual probabilities. The binomial formula $\binom{7}{x}(0.7)^x(0.3)^{7-x}$ gives each one:

• $X = 5$: there are $\binom{7}{5} = 21$ ways to choose which 5 of the 7 days are snowy.
• $X = 6$: there are $\binom{7}{6} = 7$ ways.
• $X = 7$: only one way (all seven days snowy).

Each probability is the number of arrangements times $0.7^x$ (the chance of $x$ snowy days) times $0.3^{7-x}$ (the chance of $7-x$ non-snowy days). Adding the three values gives approximately $0.647$.

Key Takeaways

• Recognise a binomial setting: fixed number of independent trials with the same success probability.
• "At least 5" means summing $P(5) + P(6) + P(7)$.
• Use $\binom{n}{x}$ for the number of ways to choose the snowy days.

Common Mistakes

• Forgetting to include one of the three terms $P(5)$, $P(6)$, or $P(7)$.
• Mixing up $p$ and $1-p$ in the formula.
• Rounding too early and losing accuracy in the final answer.

Things to Be Careful About

• The mark scheme requires the answer in the range $0.647 \leq p < 0.6475$, so keep at least three decimal places until the final step.

Walkthrough

We have already found that the probability a single week is white is $0.647$, so the probability a single week is not white is $1 - 0.647 = 0.353$.

For three independent weeks, the chance that none of them is white is the product of the individual non-white probabilities: $(0.353)^3$. This uses the multiplication rule for independent events.

"At least one is white" is the complement of "none is white", so $P(Y \geq 1) = 1 - P(Y = 0) = 1 - (0.353)^3 \approx 0.956$.

Key Takeaways

• Use the complement rule when asked about "at least one" in repeated independent trials.
• Carry the unrounded answer from part (a) into part (b) for maximum accuracy.

Common Mistakes

• Calculating $0.647^3$ instead of $(1-0.647)^3$ — the cube should apply to the non-white probability.
• Forgetting the "$1 -$" step that turns "no white weeks" into "at least one white week".

Things to Be Careful About

• The mark scheme gives an exact expression $1 - (1 - 0.647)^3$ for full marks, so show the structure of the calculation clearly.

Walkthrough

A normal approximation to a binomial is appropriate here because $n = 60$ is large and $p = 0.8$ is close to (but not too close to) 0 or 1. We first compute the binomial's mean and variance.

Step 1 — mean and variance: For $X \sim B(60, 0.8)$:

• $\mu = np = 60 \times 0.8 = 48$
• $\sigma^2 = np(1-p) = 60 \times 0.8 \times 0.2 = 9.6$, so $\sigma \approx 3.098$.

Step 2 — continuity correction: The discrete event $X < 47$ means $X \leq 46$. Under the normal approximation, this becomes the interval up to $46.5$ (halfway between 46 and 47). So we want $P(Y \leq 46.5)$.

Step 3 — standardisation: Convert to the standard normal:

Step 4 — use the table: Because the normal table typically gives $\Phi(z) = P(Z \leq z)$ for positive $z$, use the symmetry $P(Z \leq -0.4841) = 1 - P(Z \leq 0.4841) = 1 - 0.6858 = 0.3142$.

Key Takeaways

• Choose normal approximation when $n$ is large and $np$, $n(1-p)$ are reasonably big.
• Always apply a continuity correction when converting a discrete probability to a continuous one.
• Use the symmetry of the normal distribution when the standardised value is negative.

Common Mistakes

• Standardising with $\sigma^2 = 9.6$ instead of $\sigma = \sqrt{9.6}$.
• Using $47$ instead of $46.5$ (forgetting the continuity correction).
• Forgetting that "fewer than 47" means $X \leq 46$, not $X \leq 47$.
• Reading the table incorrectly for a negative $z$-value and not using symmetry.

Things to Be Careful About

• The mark scheme requires the final answer in $0.314 \leq p < 0.3145$, so be careful with the rounding of $\Phi(0.4841)$.
• The mean and variance must be computed before standardisation; if you skip straight to standardisation, you may not earn the B1 for showing the parameters.

Walkthrough

The random variable $X$ counts the number of heads when three independent biased coins are thrown. Since the coins are independent, we multiply the probability of each individual coin's outcome.

For $X=0$, there is only one way: all three coins must show tails. The tail probabilities are

so we multiply them.

For $X=1$, there are three different ways: the first coin is the only head, the second coin is the only head, or the third coin is the only head. These branches are mutually exclusive, so we add their probabilities. Each branch is the product of one head probability and two tail probabilities.

For $X=2$, similarly, there are three ways: the two tails can come from any pair of coins. Add the three products.

For $X=3$, all three coins must show heads, so multiply the three head probabilities.

Finally, check that the four probabilities add to $1$, which confirms the table is a valid probability distribution.

Key Takeaways

This question tests the multiplication law for independent events and the addition law for mutually exclusive outcomes. It also tests constructing a probability distribution table for a discrete random variable. The key idea is to break the event $X=k$ into all mutually exclusive branches, find the probability of each branch, and add them.

Common Mistakes

• Using the head probability instead of the tail probability for a tail. For example, the probability of a tail on the first coin is $\frac{2}{3}$, not $\frac{1}{3}$.
• Missing one of the three branches for $X=1$ or $X=2$. Listing the outcomes explicitly, such as $H_1T_2T_3$, $T_1H_2T_3$ and $T_1T_2H_3$, helps.
• Forgetting to check that the probabilities sum to 1.
• Trying to use a binomial model, which is not valid because the three coins have different probabilities.

Things to Be Careful About

The mark scheme accepts unsimplified probabilities, but they must be correct and identified with the correct $X$ value. The probabilities may be written as fractions with denominator 60, simplified fractions, or decimals correct to at least 3 significant figures. Ensure every probability is linked to the correct outcome in the table.

Walkthrough

For $Y=0$, all five coins must show tails. The three original coins have tail probabilities $\frac{2}{3}$, $\frac{3}{4}$ and $\frac{4}{5}$, and the two new coins each have tail probability $1-p$. Because the throws are independent, multiply these five probabilities.

For $Y=5$, all five coins must show heads. The three original coins have head probabilities $\frac{1}{3}$, $\frac{1}{4}$ and $\frac{1}{5}$, and the two new coins each have head probability $p$. Multiply again.

The condition $P(Y=0)=6P(Y=5)$ gives an equation. Multiplying through by the common denominator 60 removes the fractions and gives $24(1-p)^2=6p^2$. Dividing by 6 and expanding gives $4-8p+4p^2=p^2$, which rearranges to $3p^2-8p+4=0$. This factorises as $(3p-2)(p-2)=0$, so $p=\frac{2}{3}$ or $p=2$. Since a probability must lie between 0 and 1, $p=2$ is impossible, so the answer is $p=\frac{2}{3}$.

Key Takeaways

This question tests the multiplication law for independent events and the skill of forming and solving a quadratic equation from a probability condition. It also tests rejecting an extraneous root that is not a valid probability. The same structure appears whenever an unknown probability is found from a given probability condition.

Common Mistakes

• Using $p$ instead of $1-p$ for the tail probability of the two new coins in $P(Y=0)$.
• Forgetting to square $(1-p)$ because there are two new coins.
• Simplifying $\frac{2}{3}\times\frac{3}{4}\times\frac{4}{5}$ to $\frac{2}{5}$ too early; the mark scheme requires the unsimplified product to be shown.
• Accepting $p=2$ as a valid answer. It must be clearly rejected because a probability cannot exceed 1.
• Making arithmetic errors when expanding $4(1-p)^2$.

Things to Be Careful About

The mark scheme awards B1 for either correct expression, M1 for forming the equation $P(Y=0)=6P(Y=5)$, and A1 for the final value with $p=2$ rejected. If $p=2$ is seen and not clearly rejected, the final mark is lost. Check the factorisation: $(3p-2)(p-2)=3p^2-8p+4$.

Walkthrough

There are eight positions to fill. If all eight digits were different, the answer would be 8!. But the two 2s are identical: swapping them does not create a new arrangement, so divide by 2!. Likewise, the three 4s are identical and can be permuted among themselves in 3! ways without creating a new arrangement, so divide by 3!.

Key Takeaways

When arranging objects where some are identical, count as if all objects were different, then divide by the factorial of each repeated group.

Common Mistakes

A common error is to divide only by 3!, forgetting the repeated 2s, or not to divide at all and give 8! = 40320.

Things to Be Careful About

The denominator is the product of factorials of all repeated counts, here 2! and 3!, not their sum.

Walkthrough

With 2s fixed at both ends, the problem reduces to arranging the six middle digits. The three 4s are identical, so there are 6!/3! arrangements. We now exclude those where the three 4s are consecutive. Grouping them as a single block 444 reduces the middle to four objects: 444, 1, 3 and 5, which can be arranged in 4! ways. Subtracting gives 120 - 24 = 96.

Key Takeaways

When a condition says "not all together", count the complement (all together) and subtract it from the unrestricted count. A block of identical items counts as one item in the arrangement.

Common Mistakes

Arranging the middle as 6! without dividing by 3!; treating the block 444 as three separate items; or forgetting to subtract.

Things to Be Careful About

Because the 2s are fixed, their repetition no longer requires a 2! division in the middle count. The block subtraction uses 4!, not 4!/3!, because after grouping the 4s together they form one distinct item among four.

Walkthrough

The event "three digits are all different" means the three chosen values are distinct. Since the digit 2 has two copies and the digit 4 has three copies, a selection can include at most one 2 and at most one 4; it can never include two 2s or two 4s. The digits 1, 3, 5 each have one copy, so each contributes only one choice for its value.

We split into four exhaustive cases based on how many 2s and 4s are chosen. For each case, multiply the number of ways to pick the chosen copies by the number of ways to pick the necessary remaining digits from 1, 3, 5. Adding these gives 34 favourable selections.

Finally divide by the total number of ways to choose three physical digits from eight, which is 8C3 = 56, to obtain the probability.

Key Takeaways

To count favourable outcomes for a probability, identify repeated categories and enumerate exhaustive cases. When digits are repeated but physically distinct, the number of copies matters in combination factors.

Common Mistakes

Forgetting the factors for choosing which 2 or which 4, e.g. using 1 instead of 2 or 3; counting ordered selections with permutations instead of combinations; or using 5C3 as the total because only five different values exist.

Things to Be Careful About

Although some digits have the same value, they are still separate physical objects when counting selections, so choosing one 2 has 2 choices and choosing one 4 has 3 choices. "All different" still allows one 2 and one 4; it excludes two 2s, two 4s, or three 4s. Simplify 34/56 to 17/28.