9709/63 May/June 2024 — Mathematics Solved Past Paper

Q1Medium-EasyThe Poisson Distribution

The random variable $X$ has the distribution $\text{B}(4000, 0.001)$ .

(a)

Use a suitable approximating distribution to find $\text{P}(2 \le X < 5)$ .

DifficultyMedium-Easy

Worked solution

Approach

Since $X \sim \mathrm{B}(4000, 0.001)$ has large $n$ and small $p$ , approximate it by a Poisson distribution with $\lambda = np = 4$ . Then compute $\mathrm{P}(2 \le X < 5)$ by summing the Poisson probabilities for $X = 2, 3, 4$ .

Working

Let $X \sim \mathrm{Po}(\lambda)$ with $\lambda = 4$ . Then

\lambda = np = 4000 \times 0.001 = 4

The required probability is

\mathrm{P}(2 \le X < 5) = \mathrm{P}(X = 2) + \mathrm{P}(X = 3) + \mathrm{P}(X = 4)

Using $\mathrm{P}(X = x) = \dfrac{e^{-\lambda}\lambda^x}{x!}$ :

\mathrm{P}(2 \le X < 5) = e^{-4}\left(\frac{4^2}{2!} + \frac{4^3}{3!} + \frac{4^4}{4!}\right)

Evaluate the terms:

\frac{4^2}{2!} = 8, \quad \frac{4^3}{3!} = \frac{64}{6} = 10.6667, \quad \frac{4^4}{4!} = \frac{256}{24} = 10.6667

So

\mathrm{P}(2 \le X < 5) = e^{-4}(8 + 10.6667 + 10.6667) = e^{-4}(29.3333)

= 0.537 \quad (3\text{ sf})

Answer

\mathrm{P}(2 \le X < 5) = 0.537

Final answer

0.537

Detailed explanation

Walkthrough

We are told that $X$ has a binomial distribution with $n = 4000$ and $p = 0.001$ . Because $n$ is very large and $p$ is very small, it is sensible to approximate the binomial distribution by a Poisson distribution. The mean of the Poisson distribution is $\lambda = np = 4000 \times 0.001 = 4$ . This is why we use $\mathrm{Po}(4)$ .

We need $\mathrm{P}(2 \le X < 5)$ . Since $X$ is a discrete random variable, the condition $2 \le X < 5$ means $X = 2$ , $X = 3$ or $X = 4$ . We therefore add these three Poisson probabilities. Using $\mathrm{P}(X = x) = \dfrac{e^{-4}4^x}{x!}$ gives the expression in the solution. Evaluating the factorial terms and summing gives $0.537$ .

Key Takeaways

A binomial distribution $\mathrm{B}(n, p)$ can be approximated by a Poisson distribution when $n$ is large and $p$ is small.
The Poisson parameter is $\lambda = np$ .
A compound inequality such as $2 \le X < 5$ must be translated into the exact discrete integer values it includes.

Common Mistakes

Including $X = 5$ in $\mathrm{P}(2 \le X < 5)$ . The inequality is strict at 5, so $X = 5$ is excluded.
Calculating only one of the three probabilities instead of summing them all.
Using the binomial distribution directly, which is not the intended approach and is much more laborious.
Giving the answer without showing the Poisson expression; the mark scheme awards only partial credit for an unsupported correct answer.

Things to Be Careful About

Use the Poisson formula correctly: $\mathrm{P}(X = x) = \dfrac{e^{-\lambda}\lambda^x}{x!}$ .
Compute the factorial terms carefully: $\frac{4^3}{3!} = \frac{64}{6}$ and $\frac{4^4}{4!} = \frac{256}{24}$ .
Give the final answer to three significant figures.
In part (b), remember to justify the approximation with explicit values.

Techniques used

approximate a binomial distribution by a Poisson distributioncompute Poisson probabilities for consecutive valuessum probabilities to obtain a range probability

(b)

Justify your approximating distribution in this case.

1M Similar

DifficultyEasy

Worked solution

Approach

State the conditions that allow a binomial distribution to be approximated by a Poisson distribution: large $n$ and small $p$ .

Working

Here $n = 4000$ and $p = 0.001$ , so:

$n = 4000 > 50$
$np = 4000 \times 0.001 = 4 < 5$ (equivalently $p = 0.001 < 0.1$ )

Both conditions are satisfied, so the Poisson approximation with $\lambda = 4$ is justified.

Answer

$n = 4000 > 50$ and $np = 4 < 5$ (or $p = 0.001 < 0.1$ ).

Final answer

n = 4000 > 50 and np = 4 < 5 (or p = 0.001 < 0.1)

Detailed explanation

Walkthrough

The question asks us to justify the approximating distribution chosen in part (a). A Poisson approximation to a binomial distribution is valid when the number of trials $n$ is large and the probability of success $p$ is small. Here $n = 4000$ , which is much larger than 50, and $p = 0.001$ , which is small. The mark scheme also accepts the equivalent check $np = 4 < 5$ . We state these values explicitly to earn the mark.

Key Takeaways

The standard conditions for using the Poisson approximation to the binomial are: $n$ is large (often $n > 50$ ) and $p$ is small (often $p < 0.1$ or $np < 5$ ). The parameter of the approximating Poisson distribution is $\lambda = np$ .

Common Mistakes

Stating only the general conditions without showing the actual values $n = 4000$ and $p = 0.001$ or $np = 4$ .
Confusing the Poisson approximation conditions with those for a normal approximation. Here $np = 4 < 5$ , so a normal approximation would not be appropriate.

Things to Be Careful About

The mark scheme requires explicit values, so write $n = 4000 > 50$ and $np = 4 < 5$ (or $p = 0.001 < 0.1$ ).
Do not write only " $n$ is large and $p$ is small" without showing the numbers.

Techniques used

state the conditions for the Poisson approximationverify large n and small p with explicit values

Q2Medium-EasySampling and Estimation

The widths, $w\text{ cm}$ , of a random sample of 150 leaves of a certain kind were measured. The sample mean of $w$ was found to be $3.12\text{ cm}$ .

Using this sample, an approximate 95% confidence interval for the population mean of the widths in centimetres was found to be $[3.01, 3.23]$ .

(a)

Calculate an estimate of the population standard deviation.

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

Use the confidence interval formula

\bar{x} \pm z \frac{\sigma}{\sqrt{n}}

The interval is symmetric, so the half-width is the difference between the sample mean and either endpoint. Set this half-width equal to the margin of error and solve for $\sigma$ .

Working

For a 95% confidence interval, $z = 1.96$ . The sample mean is $\bar{x} = 3.12$ and the upper endpoint is 3.23, so the half-width is

3.23 - 3.12 = 0.11

Therefore

1.96 \times \frac{\sigma}{\sqrt{150}} = 0.11

Solving for $\sigma$ :

\sigma = \frac{0.11 \times \sqrt{150}}{1.96}

Since $\sqrt{150} = 12.247\ldots$ ,

\sigma = \frac{0.11 \times 12.247\ldots}{1.96} = 0.687\ldots

Answer

The estimate of the population standard deviation is

\sigma \approx 0.687\text{ cm} \quad (3\text{ s.f.})

Final answer

0.687 cm

Detailed explanation

Walkthrough

The confidence interval is built from the formula

\bar{x} \pm z \frac{\sigma}{\sqrt{n}}

where $n = 150$ and $z = 1.96$ for 95% confidence. The upper endpoint is $\bar{x} + z \frac{\sigma}{\sqrt{n}}$ , so the half-width is $3.23 - 3.12 = 0.11$ . Equating this to $1.96 \frac{\sigma}{\sqrt{150}}$ gives a linear equation for $\sigma$ . Solving for $\sigma$ means multiplying by $\sqrt{150}$ and dividing by $1.96$ , giving $0.687$ cm to 3 significant figures.

Key Takeaways

The confidence interval formula connects the sample mean, the standard deviation, and the z-score. Given any three of these pieces, the fourth can be found. The width of a confidence interval is twice the margin of error, so the half-width is the margin of error.

Common Mistakes

Using the full width $(3.23 - 3.01 = 0.22)$ instead of the half-width $0.11$ .
Forgetting to divide by $\sqrt{n}$ or misplacing $\sqrt{150}$ .
Using an incorrect $z$ -value, such as 1.645 or 2.576, instead of 1.96 for 95%.
Giving the standard error $\frac{\sigma}{\sqrt{n}}$ instead of the population standard deviation $\sigma$ .

Things to Be Careful About

The interval gives the standard error indirectly. Remember that the standard error is $\frac{\sigma}{\sqrt{n}}$ , so the population standard deviation is larger than the margin of error. Always include units (cm) and give the final answer to 3 significant figures as appropriate.

Techniques used

Set up confidence interval equationUse the 95% z-valueSolve for the population standard deviation

(b)

Explain whether it was necessary to use the Central Limit theorem in your answer to part (a).

1M Similar

DifficultyEasy

Worked solution

Approach

The confidence interval formula in part (a) assumes that the sample mean is approximately normally distributed. Since the population distribution of leaf widths is not given to be normal, the Central Limit Theorem is needed to justify this normality for a large sample.

Working

The population of widths is not stated as normal. For a sample of size $n = 150$ , the Central Limit Theorem states that the sample mean is approximately normally distributed, whatever the population distribution. This justifies using the normal confidence interval formula in part (a).

Answer

Yes, it was necessary to use the Central Limit Theorem, because the population distribution is not given to be normal.

Final answer

Yes, because the population is not given to be normal.

Detailed explanation

Walkthrough

The interval in part (a) used the formula $\bar{x} \pm z \frac{\sigma}{\sqrt{n}}$ , which relies on the sample mean being approximately normal. The question does not state that the widths of leaves follow a normal distribution. The Central Limit Theorem says that for a sufficiently large sample, the sample mean is approximately normal regardless of the population distribution. Since $n = 150$ is large, the CLT justifies the use of the normal confidence interval formula.

Key Takeaways

The Central Limit Theorem is essential for using normal-based confidence intervals when the population distribution is unknown or not normal. It allows us to treat the sample mean as approximately normal for large samples.

Common Mistakes

Saying "yes, because $n$ is large" without mentioning that the population is not given to be normal.
Saying "no" because the sample is large; the CLT is still necessary to justify normality.
Confusing the CLT with an assumption that the original population is normal.

Things to Be Careful About

The mark scheme requires both parts: yes, and the reason that the population distribution is not known to be normal. A statement such as "yes, because $n$ is large" is not enough on its own.

Techniques used

Recognise the need for normality of the sample meanJustify the use of the Central Limit Theorem

Q35MMediumLinear Combinations of Random Variables

The masses in kilograms of large and small bags of cement have the independent distributions $\text{N}(50, 2.4)$ and $\text{N}(26, 1.8)$ respectively.

Find the probability that the total mass of 5 randomly chosen large bags of cement is greater than the total mass of 10 randomly chosen small bags of cement.

DifficultyMedium

Worked solution

Approach

Let $L$ be the mass of a randomly chosen large bag and $S$ the mass of a randomly chosen small bag. The total mass of 5 large bags is $5L$ , and the total mass of 10 small bags is $10S$ . We need $\text{P}(5L > 10S)$ , so define the difference $D = 5L - 10S$ and find $\text{P}(D > 0)$ . Since $L$ and $S$ are independent normal variables, $D$ is also normal; we compute its mean and variance, then standardise.

Working

L \sim \text{N}(50, 2.4), \quad S \sim \text{N}(26, 1.8)

Let $D = 5L - 10S$ be the difference between the total mass of 5 large bags and the total mass of 10 small bags.

Mean of $D$ :

\text{E}(D) = 5 \times 50 - 10 \times 26 = 250 - 260 = -10

Variance of $D$ (the variances of the two independent totals add):

\text{Var}(D) = 5 \times 2.4 + 10 \times 1.8 = 12 + 18 = 30

So:

D \sim \text{N}(-10, 30)

We require $\text{P}(D > 0)$ . Standardising:

\text{P}(D > 0) = \text{P}\left(Z > \frac{0 - (-10)}{\sqrt{30}}\right) = \text{P}\left(Z > \frac{10}{\sqrt{30}}\right)

\frac{10}{\sqrt{30}} = 1.826

Therefore:

\text{P}(D > 0) = 1 - \Phi(1.826)

From the standard normal table, $\Phi(1.826) = 0.9661$ . Therefore:

\text{P}(D > 0) = 1 - 0.9661 = 0.0339

Answer

The probability that the total mass of 5 large bags exceeds the total mass of 10 small bags is:

0.0339 \ (\text{or } 0.034 \text{ to 3sf})

Final answer

0.0339

Detailed explanation

Walkthrough

We are given that a large bag has mass $L \sim \text{N}(50, 2.4)$ and a small bag has mass $S \sim \text{N}(26, 1.8)$ . We want the probability that 5 large bags weigh more than 10 small bags.

The total mass of 5 large bags is the sum of 5 independent variables, each with distribution $\text{N}(50, 2.4)$ . The mean of this total is $5 \times 50 = 250$ . Because the bags are independent, the variance of the total is $5 \times 2.4 = 12$ — the variances add, not the standard deviations.

Similarly, the total mass of 10 small bags has mean $10 \times 26 = 260$ and variance $10 \times 1.8 = 18$ .

We want $\text{P}(5L > 10S)$ , which is $\text{P}(5L - 10S > 0)$ . Define $D = 5L - 10S$ . Since $L$ and $S$ are independent normal variables, $D$ is also normal. Its mean is:

\text{E}(D) = 5 \times 50 - 10 \times 26 = 250 - 260 = -10

Its variance is:

\text{Var}(D) = 5 \times 2.4 + 10 \times 1.8 = 12 + 18 = 30

So $D \sim \text{N}(-10, 30)$ .

Now standardise: $D > 0$ is equivalent to:

Z > \frac{0 - (-10)}{\sqrt{30}} = \frac{10}{\sqrt{30}} \approx 1.826

Using the standard normal table, $\Phi(1.826) \approx 0.9661$ , so:

\text{P}(Z > 1.826) = 1 - 0.9661 = 0.0339

The key insight is that the variance of a sum of independent variables adds, and that a linear combination of independent normal variables is normal.

Key Takeaways

The mean of a linear combination of independent variables is the same linear combination of their means.
The variance of a sum of independent variables is the sum of their variances.
A linear combination of independent normal random variables is normal.
To find a probability like $\text{P}(D > 0)$ , standardise using $Z = \frac{D - \mu}{\sigma}$ and then use the standard normal table.

Common Mistakes

Using the variance of the total as $5^2 \times 2.4$ instead of $5 \times 2.4$ . The total of 5 independent bags has variance $5 \times 2.4$ , not $25 \times 2.4$ , because the coefficient of each bag in the sum is 1, not 5.
Subtracting the variances instead of adding them. Since the variables are independent, variances always add, even when one total is subtracted from another.
Getting the sign of the mean wrong: $D = 5L - 10S$ has mean $250 - 260 = -10$ , not $+10$ .
Forgetting to use $1 - \Phi(z)$ when the required probability is the upper tail.

Things to Be Careful About

Use the standard deviation $\sqrt{30}$ when standardising, not the variance 30.
The question gives variances (2.4 and 1.8), not standard deviations. The variance of the total is the sum of the variances.
No continuity correction is needed here because the normal distribution is used directly, not as an approximation to a discrete distribution.
Give the final answer to 3 significant figures: 0.0339 or 0.034.

Techniques used

compute the mean of a difference of independent normal variablescompute the variance of a difference of independent normal variablesstandardise a normal random variableread an upper-tail probability from the standard normal table

Approach

Let $X$ be the number of voters in the sample of 25 who support the Today Party. Under the null hypothesis, $X \sim B(25, 0.24)$ . We set up the hypotheses, compute the probability of observing 2 or fewer supporters, compare it with the 5% significance level, and draw a conclusion in context. Since the question says not to use an approximating distribution, we compute the exact binomial probability.

Working

State the hypotheses. Since the student wishes to test whether support has decreased, we use a one-tailed (lower-tail) test:

H_0: p = 0.24, \quad H_1: p < 0.24

The $p$ -value is the probability of observing 2 or fewer supporters under $H_0$ :

P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)

P(X \leq 2) = 0.76^{25} + 25 \times 0.76^{24} \times 0.24 + {}^{25}C_2 \times 0.76^{23} \times 0.24^2

Evaluating each term:

0.76^{25} = 0.0010479

25 \times 0.76^{24} \times 0.24 = 0.0082732

{}^{25}C_2 \times 0.76^{23} \times 0.24^2 = 300 \times 0.76^{23} \times 0.0576 = 0.0313513

Summing:

P(X \leq 2) = 0.0010479 + 0.0082732 + 0.0313513 = 0.0407

Compare with the significance level:

0.0407 < 0.05

Since the $p$ -value is less than the significance level, we reject $H_0$ .

Answer

There is sufficient evidence to suggest that support for the Today Party has decreased.

Final answer

There is sufficient evidence to suggest that support for the Today Party has decreased.

Detailed explanation

Walkthrough

This is a hypothesis test for a population proportion using the binomial distribution. The question explicitly says not to use an approximating distribution, so we must compute the exact binomial probability.

Step 1: Set up the hypotheses. The null hypothesis is that the proportion of voters supporting the Today Party is still 24%, i.e. $p = 0.24$ . The alternative hypothesis reflects what the student wants to test — that support has decreased, i.e. $p < 0.24$ . Because the alternative is "less than", this is a one-tailed (lower-tail) test.

Step 2: Identify the test statistic. Let $X$ be the number of voters in the sample supporting the Today Party. Under $H_0$ , $X \sim B(25, 0.24)$ .

Step 3: Compute the p-value. The p-value is the probability of getting a result as extreme as, or more extreme than, the observed result, assuming $H_0$ is true. Since we observed 2 supporters and the alternative is $p < 0.24$ , the p-value is $P(X \leq 2)$ . We compute this as the sum of three binomial probabilities: $P(X = 0)$ , $P(X = 1)$ , and $P(X = 2)$ . Each term uses the binomial formula $P(X = r) = {}^nC_r \, p^r (1-p)^{n-r}$ with $n = 25$ , $p = 0.24$ , and $1-p = 0.76$ .

Step 4: Compare with the significance level. The significance level is 5%, i.e. 0.05. Since $0.0407 < 0.05$ , the result is significant — the observed data is unlikely under the null hypothesis.

Step 5: Draw a conclusion. Because the p-value is less than the significance level, we reject $H_0$ and conclude that there is sufficient evidence to suggest support for the Today Party has decreased.

Key Takeaways

A hypothesis test for a proportion uses the binomial distribution $B(n, p)$ when the sample is small and the question requires exact computation.
The alternative hypothesis determines the direction of the test: $p < p_0$ gives a one-tailed lower-tail test.
The p-value is the probability of observing a result as extreme as, or more extreme than, the one observed, under the null hypothesis.
We reject $H_0$ when the p-value is less than the significance level.
The conclusion must be stated in context and should not be over-certain ("there is evidence to suggest" rather than "it is certain").

Common Mistakes

Not showing the probability expression. The mark scheme awards M1 only if the expression for $P(X \leq 2)$ is seen. An unsupported answer of 0.0407 earns only SC B1.
Using a two-tailed test. If the student sets $H_1: p \neq 0.24$ and compares with 0.025, the mark scheme caps the marks at B0 M1 A1 M1 A0.
Using an approximating distribution. The question explicitly says not to use an approximating distribution, so a normal approximation would not be appropriate.
Not stating the conclusion in context. The final conclusion must mention the Today Party and the decrease in support.

Things to Be Careful About

The hypotheses must be exactly $H_0: p = 0.24$ and $H_1: p < 0.24$ .
The probability must be computed exactly using the binomial formula, not approximated.
The comparison must be with 0.05, the significance level, not with 0.025 (which would be for a two-tailed test).
The conclusion should be phrased as "sufficient evidence to suggest that support has decreased" — not as a definite statement, and without contradictions.

Techniques used

define null and alternative hypothesescompute exact binomial tail probabilitycompare p-value with significance leveldraw conclusion in context

Approach

For a probability density function, the total area under the curve over its support must equal $1$ . Integrate $f(x)$ from $0$ to $\sqrt{2}$ , set the result equal to $1$ , and solve for $a$ .

Working

\int_0^{\sqrt{2}} (ax - x^3)\,dx = 1

Integrating:

\left[\frac{ax^2}{2} - \frac{x^4}{4}\right]_0^{\sqrt{2}} = 1

Substituting the limits:

\frac{a(\sqrt{2})^2}{2} - \frac{(\sqrt{2})^4}{4} = 1

\frac{2a}{2} - \frac{4}{4} = 1

a - 1 = 1

a = 2

Answer

a = 2

Final answer

a = 2

Detailed explanation

Walkthrough

The defining property of a probability density function is that the total area under the curve over its entire support equals $1$ . Since $f(x) = 0$ outside the interval $[0, \sqrt{2}]$ , the total probability is just the integral from $0$ to $\sqrt{2}$ of $f(x)$ . We therefore set

\int_0^{\sqrt{2}} (ax - x^3)\,dx = 1.

We integrate term by term: the antiderivative of $ax$ is $ax^2/2$ and the antiderivative of $x^3$ is $x^4/4$ . Substituting the upper limit $\sqrt{2}$ :

$a(\sqrt{2})^2/2 = a \cdot 2/2 = a$
$(\sqrt{2})^4/4 = 4/4 = 1$

The lower limit $0$ contributes $0$ to both terms. So the integral evaluates to $a - 1$ . Setting this equal to $1$ gives $a - 1 = 1$ , hence $a = 2$ . This confirms the given value.

Key Takeaways

This question tests the fundamental property of a probability density function: the total area under the curve must equal 1. It also tests the ability to integrate a polynomial and evaluate it at the limits of integration. The key skill is translating the PDF property into a definite integral equation.

Common Mistakes

Forgetting to set the integral equal to 1 — the total probability must be 1.
Arithmetic errors with surds: $(\sqrt{2})^2 = 2$ and $(\sqrt{2})^4 = 4$ .
Forgetting to state that the lower limit contributes 0 (though here it does, it should still be acknowledged).
Incorrectly integrating $ax - x^3$ (e.g. dividing by the wrong power).

Things to Be Careful About

The support of the PDF is $[0, \sqrt{2}]$ , so the integral must be taken over exactly this interval. Since $f(x) = 0$ elsewhere, no other intervals contribute. Also note that the question says "Show that $a = 2$ " — the mark scheme requires the result to be convincingly obtained with no errors, so every step of the integration and substitution must be shown.

Techniques used

integrate the PDF over its supportequate total probability to 1substitute limits and solve for the constant

(b)

Find the median of $X$ .

4M Similar

DifficultyMedium

Worked solution

Approach

The median $m$ satisfies $\int_0^m f(x)\,dx = \frac{1}{2}$ . With $a = 2$ , integrate $2x - x^3$ from $0$ to $m$ , set the result equal to $\frac{1}{2}$ , and solve the resulting quartic by substituting $u = m^2$ .

Working

The median $m$ satisfies:

\int_0^m (2x - x^3)\,dx = \frac{1}{2}

Integrating:

\left[x^2 - \frac{x^4}{4}\right]_0^m = \frac{1}{2}

m^2 - \frac{m^4}{4} = \frac{1}{2}

Multiplying by 4:

4m^2 - m^4 = 2

m^4 - 4m^2 + 2 = 0

Let $u = m^2$ . Then:

u^2 - 4u + 2 = 0

u = \frac{4 \pm \sqrt{16 - 8}}{2} = \frac{4 \pm \sqrt{8}}{2} = 2 \pm \sqrt{2}

Since $0 \le m \le \sqrt{2}$ , we need $m^2 \le 2$ . The value $2 + \sqrt{2} \approx 3.414 > 2$ is invalid, so:

m^2 = 2 - \sqrt{2}

m = \sqrt{2 - \sqrt{2}} \approx 0.765

Answer

m = \sqrt{2 - \sqrt{2}} \approx 0.765

Final answer

m = sqrt(2 - sqrt(2)) ≈ 0.765

Detailed explanation

Walkthrough

The median $m$ of a continuous random variable is the value such that $P(X \le m) = \frac{1}{2}$ . Since the support of $X$ starts at $0$ , this means

\int_0^m f(x)\,dx = \frac{1}{2}.

With $a = 2$ from part (a), the PDF is $f(x) = 2x - x^3$ . Integrating from $0$ to $m$ :

\left[x^2 - \frac{x^4}{4}\right]_0^m = m^2 - \frac{m^4}{4}.

Setting this equal to $\frac{1}{2}$ and multiplying through by 4 gives the quartic $m^4 - 4m^2 + 2 = 0$ . This is a quartic that is quadratic in $m^2$ , so we substitute $u = m^2$ to obtain $u^2 - 4u + 2 = 0$ . The quadratic formula gives

u = \frac{4 \pm \sqrt{16 - 8}}{2} = 2 \pm \sqrt{2}.

Now we must select the valid root. Since $m$ lies in the support $[0, \sqrt{2}]$ , we need $m^2 \le 2$ . The value $2 + \sqrt{2} \approx 3.414$ exceeds 2 and is therefore invalid. Hence $m^2 = 2 - \sqrt{2}$ and $m = \sqrt{2 - \sqrt{2}} \approx 0.765$ .

Key Takeaways

This question combines the definition of the median of a continuous distribution with solving a quartic equation by substitution. The key steps are: recognising that the median satisfies an integral equation, integrating correctly, reducing the quartic to a quadratic via $u = m^2$ , and rejecting the root that lies outside the support.

Common Mistakes

Forgetting the $\frac{1}{2}$ on the right-hand side of the median equation.
Not rejecting the root $m^2 = 2 + \sqrt{2}$ , which gives $m \approx 1.848$ , outside the support $[0, \sqrt{2}]$ .
Errors in the quadratic formula, especially with $\sqrt{8} = 2\sqrt{2}$ .
Taking the negative square root of $m^2$ — the median must be positive since the support is $[0, \sqrt{2}]$ .

Things to Be Careful About

Always check that the median lies within the support of the distribution. Here the support is $[0, \sqrt{2}] \approx [0, 1.414]$ , and $\sqrt{2 - \sqrt{2}} \approx 0.765$ is comfortably inside. The quartic has four possible roots for $m$ (two for $m^2$ , each with $\pm$ ), but only the positive root of $m^2 = 2 - \sqrt{2}$ is valid.

Techniques used

set up the median equationintegrate the PDF from 0 to msolve the quartic by substitutionselect the root within the support

(c)

Find the exact value of $\text{E}(X)$ .

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

Use the definition $\text{E}(X) = \int x f(x)\,dx$ over the support of $X$ . With $a = 2$ , this becomes $\int_0^{\sqrt{2}} (2x^2 - x^4)\,dx$ .

Working

\text{E}(X) = \int_0^{\sqrt{2}} x(2x - x^3)\,dx = \int_0^{\sqrt{2}} (2x^2 - x^4)\,dx

\text{E}(X) = \left[\frac{2x^3}{3} - \frac{x^5}{5}\right]_0^{\sqrt{2}}

\text{E}(X) = \frac{2(\sqrt{2})^3}{3} - \frac{(\sqrt{2})^5}{5}

Since $(\sqrt{2})^3 = 2\sqrt{2}$ and $(\sqrt{2})^5 = 4\sqrt{2}$ :

\text{E}(X) = \frac{4\sqrt{2}}{3} - \frac{4\sqrt{2}}{5}

\text{E}(X) = 4\sqrt{2}\left(\frac{1}{3} - \frac{1}{5}\right) = 4\sqrt{2} \cdot \frac{2}{15} = \frac{8\sqrt{2}}{15}

Answer

\text{E}(X) = \frac{8\sqrt{2}}{15}

Final answer

E(X) = 8√2/15

Detailed explanation

Walkthrough

The expected value of a continuous random variable is defined as

\text{E}(X) = \int_{-\infty}^{\infty} x f(x)\,dx.

Since $f(x) = 0$ outside $[0, \sqrt{2}]$ , the integral reduces to $\int_0^{\sqrt{2}} x f(x)\,dx$ . With $f(x) = 2x - x^3$ , we multiply by $x$ to obtain $2x^2 - x^4$ . Integrating:

\int_0^{\sqrt{2}} (2x^2 - x^4)\,dx = \left[\frac{2x^3}{3} - \frac{x^5}{5}\right]_0^{\sqrt{2}}.

At the upper limit $\sqrt{2}$ :

$2(\sqrt{2})^3/3 = 2(2\sqrt{2})/3 = 4\sqrt{2}/3$
$(\sqrt{2})^5/5 = 4\sqrt{2}/5$

The lower limit $0$ contributes $0$ . So

\text{E}(X) = \frac{4\sqrt{2}}{3} - \frac{4\sqrt{2}}{5} = 4\sqrt{2}\left(\frac{1}{3} - \frac{1}{5}\right) = 4\sqrt{2} \cdot \frac{2}{15} = \frac{8\sqrt{2}}{15}.

Key Takeaways

This question tests the definition of the expected value of a continuous random variable: $\text{E}(X) = \int x f(x)\,dx$ . The key skills are multiplying the PDF by $x$ before integrating, integrating the resulting polynomial, and simplifying surds such as $(\sqrt{2})^3 = 2\sqrt{2}$ and $(\sqrt{2})^5 = 4\sqrt{2}$ .

Common Mistakes

Forgetting to multiply $f(x)$ by $x$ before integrating — a very common error.
Arithmetic errors with surd powers: $(\sqrt{2})^3 = 2\sqrt{2}$ , $(\sqrt{2})^5 = 4\sqrt{2}$ .
Incorrectly combining the fractions $\frac{4\sqrt{2}}{3}$ and $\frac{4\sqrt{2}}{5}$ .
Using the wrong limits (e.g. integrating from $0$ to $m$ rather than $0$ to $\sqrt{2}$ ).

Things to Be Careful About

The integral for $\text{E}(X)$ must use $x \cdot f(x)$ , not $f(x)$ alone. Also, the limits are the full support $[0, \sqrt{2}]$ , not the median from part (b). Finally, the answer must be given as a single exact term, $\frac{8\sqrt{2}}{15}$ , as required by the mark scheme.

Techniques used

form the expectation integralmultiply the PDF by xintegrate and evaluate at the limits

Q6MediumSampling and EstimationHypothesis Tests

The numbers of green sweets in 200 randomly chosen packets of Frutos are summarised in the table.

Number of green sweets	0	1	2	3	$> 3$
Number of packets	32	50	97	21	0

(a)

Calculate an unbiased estimate for the population mean of the number of green sweets in a packet of Frutos, and show that an unbiased estimate of the population variance is 0.783 correct to 3 significant figures.

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

Estimate the population mean by the sample mean and estimate the population variance using the unbiased formula with $n-1$ in the divisor. Since the data are grouped, use the sums $\sum x f$ and $\sum x^2 f$ .

Working

The total number of packets is

n = 32 + 50 + 97 + 21 + 0 = 200.

The sum of the observed values is

\sum x f = 0(32) + 1(50) + 2(97) + 3(21) = 50 + 194 + 63 = 307.

Therefore the unbiased estimate of the population mean is

\hat{\mu} = \bar{x} = \frac{307}{200} = 1.535.

The sum of squares is

\sum x^2 f = 0^2(32) + 1^2(50) + 2^2(97) + 3^2(21) = 50 + 388 + 189 = 627.

The unbiased estimate of the population variance is

s^2 = \frac{1}{n-1}\left(\sum x^2 f - \frac{(\sum x f)^2}{n}\right)

= \frac{1}{199}\left(627 - \frac{307^2}{200}\right)

= \frac{1}{199}(627 - 471.245) = \frac{155.755}{199} = 0.7827\ldots

= 0.783 \quad \text{to 3 s.f.}

Answer

$\hat{\mu} = 1.535$ , and the unbiased estimate of the population variance is $0.783$ to 3 significant figures.

Final answer

mean estimate = 1.535, unbiased variance estimate = 0.783 (3 s.f.)

Detailed explanation

Walkthrough

The table gives the frequencies of packets containing 0, 1, 2 or 3 green sweets. The category $>3$ has frequency 0, so it contributes nothing. First find the total number of packets, $n=200$ . Then compute the weighted sum $\sum x f$ , where each number of sweets is multiplied by its frequency. Dividing by $n$ gives the sample mean, which is an unbiased estimator of the population mean.

For the variance, use the grouped-data unbiased variance formula. This uses $\sum x^2 f$ and subtracts $n\bar{x}^2$ , or equivalently $(\sum x f)^2/n$ , then divides by $n-1$ instead of $n$ . The $n-1$ divisor is what makes the estimate unbiased for the population variance.

Key Takeaways

The sample mean $\bar{x}$ is an unbiased estimate of the population mean $\mu$ . The unbiased variance estimate is

s^2 = \frac{1}{n-1}\left(\sum x^2 f - \frac{(\sum x f)^2}{n}\right).

Grouped frequency tables require weighted sums using the frequencies.

Common Mistakes

Using $n$ instead of $n-1$ in the variance denominator. Forgetting that the $0$ category contributes nothing to $\sum x f$ and $\sum x^2 f$ . Rounding the mean to 1.54 or 1.53 before calculating the variance, which loses accuracy.

Things to Be Careful About

The $>3$ category has frequency 0, so it is ignored. Use the exact mean $1.535$ in the variance calculation. The final variance must be shown as $0.783$ to 3 significant figures, with the working clearly leading to that value.

Techniques used

compute the sample mean from grouped datacompute the sum of squared valuesapply the unbiased variance formula with n-1

(b)

The manufacturers of Frutos claim that the mean number of green sweets in a packet is 1.65.

Anji believes that the true value of the mean, $\mu$ , is less than 1.65. She uses the results from the 200 randomly chosen packets to test the manufacturers' claim.

State suitable null and alternative hypotheses for the test.

1M Similar

(c)

Show that the result of Anji's test is significant at the 5% level but not at the 1% level.

4M Similar

DifficultyMedium

Worked solution

Approach

Since the sample size is large, the sample mean is approximately normal. Use the sample mean and the unbiased variance estimate from part (a), standardise to find the test statistic, and compare it with the lower-tail critical values for the 5% and 1% levels.

Working

From part (a), the sample mean and variance estimate are

\bar{x} = 1.535, \qquad s^2 = 0.783, \qquad n = 200.

The test statistic is

z = \frac{\bar{x} - \mu_0}{\sqrt{s^2/n}} = \frac{1.535 - 1.65}{\sqrt{0.783/200}}.

Now

\sqrt{0.783/200} = \sqrt{0.003915} = 0.06257,

so

z = \frac{-0.115}{0.06257} = -1.838.

The lower-tail critical values are

z_{0.05} = -1.645, \qquad z_{0.01} = -2.326.

Comparing:

-2.326 < -1.838 < -1.645.

Since $z < -1.645$ , $H_0$ is rejected at the 5% level. Since $z > -2.326$ , $H_0$ is not rejected at the 1% level.

Equivalently, the p-value is

P(Z < -1.838) = 0.033,

and

0.01 < 0.033 < 0.05.

Answer

The result is significant at the 5% level but not at the 1% level.

Final answer

Significant at 5% but not at 1%

Detailed explanation

Walkthrough

The sample size is 200, so the Central Limit Theorem tells us the sample mean is approximately normal. We estimate the population variance by $s^2 = 0.783$ from part (a). The test statistic measures how many standard errors the sample mean is below the claimed mean 1.65.

The denominator is the standard error of the sample mean, $\sqrt{s^2/n}$ , not just $\sqrt{s^2}$ . After standardising, we get $z = -1.838$ .

For a one-tailed test at the 5% level, the critical value is $-1.645$ . Because $-1.838$ is less than $-1.645$ , the result lies in the 5% rejection region. At the 1% level, the critical value is $-2.326$ . Because $-1.838$ is greater than $-2.326$ , it is not in the 1% rejection region.

The same conclusion comes from the p-value: $0.033$ is less than $0.05$ but greater than $0.01$ .

Key Takeaways

In a hypothesis test for a population mean with a large sample, use $z = \frac{\bar{x} - \mu_0}{\sqrt{s^2/n}}$ . Compare the test statistic with the appropriate one-tailed critical values, or compare the p-value with the significance level.

Common Mistakes

Using two-tailed critical values $\pm 1.96$ and $\pm 2.576$ instead of one-tailed values $-1.645$ and $-2.326$ . Forgetting to divide $s^2$ by $n$ in the denominator. Using 1.54 or 1.53 instead of 1.535, which can change the exact value of the test statistic.

Things to Be Careful About

The signs matter: the test statistic is negative, and the critical values are negative. $-1.838$ is less than $-1.645$ but greater than $-2.326$ . The p-value must be between 0.01 and 0.05 for the required conclusion.

Techniques used

standardise the sample meancompare the test statistic with one-tailed critical valuesinterpret the p-value

(d)

It is given that Anji made a Type I error.

Explain how this shows that the significance level that Anji used in her test was not 1%.

1M Similar

Q7MediumLinear Combinations of Random VariablesThe Poisson DistributionSampling and Estimation

The independent random variables $X$ and $Y$ have the distributions $\text{Po}(1.9)$ and $\text{Po}(2.2)$ respectively.

(a)

Find $\text{P}(X + Y < 4)$ .

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

Since $X$ and $Y$ are independent Poisson random variables, their sum $X + Y$ is also Poisson with parameter $\lambda = 1.9 + 2.2 = 4.1$ . We calculate $\text{P}(X+Y < 4)$ by summing the Poisson probabilities for $X+Y = 0, 1, 2, 3$ .

Working

Since $X \sim \text{Po}(1.9)$ and $Y \sim \text{Po}(2.2)$ are independent:

X + Y \sim \text{Po}(4.1)

We need:

\text{P}(X+Y < 4) = \text{P}(X+Y=0) + \text{P}(X+Y=1) + \text{P}(X+Y=2) + \text{P}(X+Y=3)

Using $\text{P}(W=k) = \dfrac{e^{-\lambda}\lambda^k}{k!}$ with $\lambda = 4.1$ :

\text{P}(X+Y=0) = e^{-4.1} = 0.01657

\text{P}(X+Y=1) = e^{-4.1} \times 4.1 = 0.06795

\text{P}(X+Y=2) = e^{-4.1} \times \frac{4.1^2}{2!} = e^{-4.1} \times 8.405 = 0.13929

\text{P}(X+Y=3) = e^{-4.1} \times \frac{4.1^3}{3!} = e^{-4.1} \times 11.4868 = 0.19037

Summing the four probabilities:

\text{P}(X+Y<4) = 0.01657 + 0.06795 + 0.13929 + 0.19037 = 0.41418

Answer

\text{P}(X+Y < 4) = 0.414 \quad (3\text{ sf})

Final answer

0.414

Detailed explanation

Walkthrough

The key insight is that when two independent Poisson variables are added, the result is also Poisson with a parameter equal to the sum of the parameters. Here, $X \sim \text{Po}(1.9)$ and $Y \sim \text{Po}(2.2)$ are independent, so $X+Y \sim \text{Po}(4.1)$ .

The event $X+Y < 4$ means $X+Y$ takes values $0, 1, 2,$ or $3$ . Since these are mutually exclusive events, we add their individual probabilities. Each probability comes from the Poisson formula $\text{P}(W=k) = e^{-\lambda}\lambda^k/k!$ with $\lambda = 4.1$ .

We compute each term: $e^{-4.1}$ for $k=0$ , $e^{-4.1} \times 4.1$ for $k=1$ , $e^{-4.1} \times 4.1^2/2!$ for $k=2$ , and $e^{-4.1} \times 4.1^3/3!$ for $k=3$ . Summing these four terms gives approximately 0.414.

Key Takeaways

The sum of independent Poisson random variables is Poisson with parameter equal to the sum of the parameters.
To find $\text{P}(W < k)$ for a Poisson variable, sum the individual probabilities $\text{P}(W=0) + \text{P}(W=1) + \cdots + \text{P}(W=k-1)$ .
The Poisson probability formula $\text{P}(W=k) = e^{-\lambda}\lambda^k/k!$ is the fundamental tool.

Common Mistakes

Forgetting that $X+Y$ is Poisson — trying to compute the probability directly from $X$ and $Y$ separately.
Using the wrong parameter (e.g., using 1.9 or 2.2 instead of 4.1).
Including $\text{P}(X+Y=4)$ in the sum — the inequality is strict ( $< 4$ ), so we stop at $k=3$ .
Arithmetic errors in computing powers and factorials.

Things to Be Careful About

The strict inequality $X+Y < 4$ means we include $k = 0, 1, 2, 3$ but NOT $k = 4$ .
Keep sufficient precision in intermediate steps; the final answer should be given to 3 significant figures.
The mark scheme allows "one end error" — but you should aim for the correct range.

Techniques used

use the sum of independent Poisson variables propertycalculate Poisson probabilities for each value of the sumsum individual probabilities to find a cumulative probability

(b)

Find the probability that $X = 2$ given that $X + Y < 4$ .

4M Similar

DifficultyMedium

Worked solution

Approach

Use the conditional probability formula:

\text{P}(X=2 \mid X+Y<4) = \frac{\text{P}(X=2 \text{ and } X+Y<4)}{\text{P}(X+Y<4)}

The denominator is the result from part (a), $\text{P}(X+Y<4) = 0.414$ . For the numerator, when $X=2$ and $X+Y<4$ , we need $Y=0$ or $Y=1$ .

Working

The joint event $X=2$ and $X+Y<4$ occurs when $X=2$ and $Y=0$ or $Y=1$ :

\text{P}(X=2 \text{ and } X+Y<4) = \text{P}(X=2, Y=0) + \text{P}(X=2, Y=1)

Since $X$ and $Y$ are independent:

\text{P}(X=2, Y=0) = \text{P}(X=2) \times \text{P}(Y=0) = e^{-1.9}\frac{1.9^2}{2!} \times e^{-2.2}

\text{P}(X=2, Y=1) = \text{P}(X=2) \times \text{P}(Y=1) = e^{-1.9}\frac{1.9^2}{2!} \times e^{-2.2} \times 2.2

Therefore:

\text{P}(X=2 \text{ and } X+Y<4) = e^{-1.9} \times \frac{1.9^2}{2} \times e^{-2.2} \times (1 + 2.2)

= 0.14957 \times 1.805 \times 0.11080 \times 3.2 = 0.09571

Now apply the conditional probability formula:

\text{P}(X=2 \mid X+Y<4) = \frac{0.09571}{0.414} = 0.23116

Answer

\text{P}(X=2 \mid X+Y<4) = 0.231 \quad (3\text{ sf})

Final answer

0.231

Detailed explanation

Walkthrough

This part uses the conditional probability formula $\text{P}(A \mid B) = \text{P}(A \cap B) / \text{P}(B)$ . Here $A$ is the event $X=2$ and $B$ is the event $X+Y<4$ .

The denominator $\text{P}(X+Y<4) = 0.414$ was found in part (a).

For the numerator, we need the joint event $X=2$ AND $X+Y<4$ . When $X=2$ , the condition $X+Y<4$ means $Y<2$ , so $Y=0$ or $Y=1$ . Since $X$ and $Y$ are independent, we can multiply their individual Poisson probabilities.

$\text{P}(X=2) = e^{-1.9} \times 1.9^2/2!$ , $\text{P}(Y=0) = e^{-2.2}$ , and $\text{P}(Y=1) = e^{-2.2} \times 2.2$ .

We compute $\text{P}(X=2, Y=0) + \text{P}(X=2, Y=1) = 0.0957$ , then divide by 0.414 to get the conditional probability 0.231.

Key Takeaways

Conditional probability: $\text{P}(A \mid B) = \text{P}(A \cap B) / \text{P}(B)$ .
For independent variables, joint probabilities factor into products of marginal probabilities.
The condition "given $X+Y<4$ " restricts the possible values of $Y$ once $X$ is fixed.

Common Mistakes

Forgetting to multiply by $\text{P}(Y=0)$ or $\text{P}(Y=1)$ — the joint probability is a product, not just $\text{P}(X=2)$ .
Using the wrong range for $Y$ (e.g., including $Y=2$ ).
Forgetting to divide by $\text{P}(X+Y<4)$ — giving the joint probability instead of the conditional probability.
Using the wrong value for $\text{P}(X+Y<4)$ from part (a).

Things to Be Careful About

The numerator is the JOINT probability $\text{P}(X=2 \text{ and } X+Y<4)$ , not $\text{P}(X=2)$ .
Make sure both $Y=0$ and $Y=1$ cases are included.
The mark scheme requires the division step to be shown explicitly.

Techniques used

apply the conditional probability formulacalculate joint Poisson probabilities for independent variablesdivide the joint probability by the total probability

(c)

A sample of 60 randomly chosen pairs of values of $X$ and $Y$ is taken, and the value of $X + Y$ is calculated for each pair. The sample mean of these 60 values is found.

Find the probability that the sample mean of $X + Y$ is less than 4.0.

6M Similar

DifficultyMedium

Worked solution

Approach

From the Poisson sum property, $W = X+Y \sim \text{Po}(4.1)$ , so $\text{E}(W) = 4.1$ and $\text{Var}(W) = 4.1$ . For a sample of 60 values, the sample mean $\bar{W}$ is approximately normally distributed by the Central Limit Theorem with mean $4.1$ and variance $4.1/60$ . Standardise and find the required probability.

Working

For $W = X+Y \sim \text{Po}(4.1)$ :

\text{E}(W) = 4.1, \quad \text{Var}(W) = 4.1

For a sample of $n = 60$ values, the sample mean is approximately normal:

\bar{W} \sim \text{N}\left(4.1, \frac{4.1}{60}\right)

Standardise:

\text{P}(\bar{W} < 4.0) = \text{P}\left(Z < \frac{4.0 - 4.1}{\sqrt{4.1/60}}\right)

\sqrt{\frac{4.1}{60}} = \sqrt{0.068333} = 0.26141

Z = \frac{-0.1}{0.26141} = -0.3826 \approx -0.383

Using the symmetry of the standard normal distribution:

\text{P}(Z < -0.383) = \Phi(-0.383) = 1 - \Phi(0.383)

From standard normal tables, $\Phi(0.383) = 0.649$ :

1 - 0.649 = 0.351

Answer

\text{P}(\text{sample mean} < 4.0) = 0.351 \quad (3\text{ sf})

Note: An alternative totals method uses the sum of the 60 values, $T \sim \text{N}(246, 246)$ approximately, giving $\text{P}(T < 240) = \text{P}(Z < (240-246)/\sqrt{246}) = \text{P}(Z < -0.383) = 0.351$ . A continuity correction would adjust this to approximately 0.340.

Final answer

0.351

Detailed explanation

Walkthrough

This part applies the Central Limit Theorem. We know $W = X+Y \sim \text{Po}(4.1)$ , so $\text{E}(W) = 4.1$ and $\text{Var}(W) = 4.1$ (for a Poisson distribution, mean = variance = $\lambda$ ).

We take a sample of 60 values of $W$ . The sample mean $\bar{W}$ has mean $\text{E}(\bar{W}) = 4.1$ and variance $\text{Var}(\bar{W}) = 4.1/60$ . By the Central Limit Theorem, for a large sample ( $n = 60$ ), $\bar{W}$ is approximately normally distributed.

We standardise: $Z = (\bar{W} - 4.1)/\sqrt{4.1/60}$ . Then $\text{P}(\bar{W} < 4.0) = \text{P}(Z < (4.0-4.1)/\sqrt{4.1/60})$ .

Computing: $\sqrt{4.1/60} = 0.2614$ , so $Z = -0.1/0.2614 = -0.383$ .

Since the standard normal distribution is symmetric, $\Phi(-0.383) = 1 - \Phi(0.383)$ . From tables, $\Phi(0.383) = 0.649$ , so the probability is $1 - 0.649 = 0.351$ .

Key Takeaways

The Central Limit Theorem: for large samples, the sample mean is approximately normal regardless of the population distribution.
For a sample mean: $\text{E}(\bar{X}) = \mu$ and $\text{Var}(\bar{X}) = \sigma^2/n$ .
For a Poisson distribution, the mean and variance are both equal to $\lambda$ .
Standardising: $Z = (\bar{X} - \mu)/\sqrt{\sigma^2/n}$ .

Common Mistakes

Using the variance of $W$ directly instead of dividing by $n$ — the sample mean has variance $\sigma^2/n$ , not $\sigma^2$ .
Forgetting to take the square root of the variance when standardising.
Using $\text{P}(Z > z)$ instead of $\text{P}(Z < z)$ — check the direction of the inequality.
Mixing up the totals method and the sample mean method.

Things to Be Careful About

The CLT applies because $n = 60$ is a large sample.
The mean and variance of $X+Y$ must come from the Poisson sum property.
If using the totals method, the sum of 60 values has mean $60 \times 4.1 = 246$ and variance $60 \times 4.1 = 246$ .
A continuity correction is sometimes applied (giving approximately 0.340), but the standard answer without it is 0.351. The mark scheme accepts both.

Techniques used

apply the Central Limit Theorem to the sample meanstandardise the sample mean using its mean and varianceuse the standard normal distribution to find the probability

Mathematics 9709/63 — May/June 2024

Approach

Working

Answer

Walkthrough

Key Takeaways

Common Mistakes

Things to Be Careful About

Approach

Working

Answer

Walkthrough

Key Takeaways

Common Mistakes

Things to Be Careful About

Approach

Working

Answer

Walkthrough

Key Takeaways

Common Mistakes

Things to Be Careful About

Approach

Working

Answer

Walkthrough

Key Takeaways

Common Mistakes

Things to Be Careful About

Approach

Working

Answer

Walkthrough

Key Takeaways

Common Mistakes

Things to Be Careful About

Approach

Working

Answer

Walkthrough

Key Takeaways

Common Mistakes

Things to Be Careful About

Approach

Working

Answer

Walkthrough

Key Takeaways

Common Mistakes

Things to Be Careful About

Approach

Working

Answer

Walkthrough

Key Takeaways

Common Mistakes

Things to Be Careful About

Approach

Working

Answer

Walkthrough

Key Takeaways

Common Mistakes

Things to Be Careful About

Approach

Working

Answer

Walkthrough

Key Takeaways

Common Mistakes

Things to Be Careful About

Approach

Working

Answer

Walkthrough

Key Takeaways

Common Mistakes

Things to Be Careful About

Approach

Working