CAIE A-Level Further Math AS 4 Further Probability and Statistics Questions

Practise Further Probability and Statistics through distributions, inference, chi-squared tests and generating functions, with mark schemes for every calculation.

Covered syllabus points

4.1 Continuous random variables 4.2 Inference using normal and t-distributions 4.3 Chi-squared tests 4.4 Non-parametric tests 4.5 Probability generating functions

[Maximum number: 9]

Figure for Question 1 — CAIE A-Level Further Math AS

As shown in the diagram, the continuous random variable X has probability density function f given by

f(x)= \begin{cases}m x & 0 \leqslant x \leqslant 2 \\ \frac{k}{x^{2}}+c & 2 \leqslant x \leqslant 6 \\ 0 & \text { otherwise }\end{cases}

where m, k and c are constants.

(a)

Given that $\mathrm{P}(X \leqslant 2)=\frac{1}{3}$ , show that $m=\frac{1}{6}$ and find the values of k and c.

[ 4 ]

$\begin{aligned} \mathrm{P}(X \leqslant 2)=\frac{1}{3}, \quad \text { so }\left[\frac{m x^{2}}{2}\right]_{0}^{2}=\frac{1}{3} m\left(\frac{4}{2}\right)=\frac{1}{3}, \quad m=\frac{1}{6} \end{aligned}$
AG
Reasoning required.

$\int_{2}^{6}\left(\frac{k}{x^{2}}+c\right) \mathrm{d} x=\frac{2}{3} \quad$ so $\quad\left[-\frac{k}{x}+c x\right]_{2}^{6}=4 c+\frac{1}{3} k=\frac{2}{3} \quad$ or $\quad k+12 c=2$
Attempt to integrate and correct use of correct limits to
form linear equation in k and c.

Also, $2 m=\frac{k}{4}+c$ or $\frac{1}{3}=\frac{k}{4}+c$ or 3 k+12 c=4
Matching at x=2.

Solve, $k=1, \quad c=\frac{1}{12}$
4

(b)

Find the exact numerical value of the interquartile range of X.

[ 5 ]

LQ: $\int_{0}^{L} \frac{1}{6} x \mathrm{~d} x=\frac{1}{4}, \quad L=\sqrt{3}$
UQ: $\frac{1}{3}+\int_{2}^{U}\left(\frac{k}{x^{2}}+c\right) \mathrm{d} x=\frac{3}{4}$
Or find and use CDF. May be in terms of m.

$\begin{aligned} -\frac{k}{U}+U c+\frac{k}{2}-2 c=\frac{5}{12} U^{2}-U-12=0 \end{aligned}$
Attempt at integral and correct use of correct limits.
Simplify to quadratic equation, may be in terms of k and
c.

U=4
$\mathrm{IQR}=4-\sqrt{3}$

A1 FT

FT their UQ - their LQ, exact values only.

5

[Maximum number: 6]

Maya is an athlete who competes in 1500 -metre races. Last summer her practice run times had mean 4.22 minutes. Over the winter she has done some intense training to try to improve her times. A random sample of 10 of her practice run times, x minutes, this summer are summarised as follows.

\sum x=42.05 \quad \sum x^{2}=176.83

Maya's new practice run times are normally distributed. She believes that on average her times have improved as a result of her training.

Test, at the 5% significance level, whether Maya's belief is supported by the data.

$\mathrm{H}_{0}: \mu=4.22 \quad \mathrm{H}_{1}: \mu<4.22$ $\bar{x}=\frac{42.05}{10}[=4.205], \quad s^{2}=\frac{1}{9}\left(176.83-\frac{42.05^{2}}{10}\right)\left[=\frac{13}{12000}=0.0010833\right]$ $t=\frac{4.205-4.22}{\sqrt{\frac{s^{2}}{10}}}$
Their 4.205, their $s^{2}$ .

t=-1.44
Condone sign.

Tabular value =1.833: ' 1.44 ' <1.833, accept $\mathrm{H}_{0}$
Compare their value with correct tabular value, signs
consistent, allow 'not significant'.

There is insufficient evidence to support Maya's belief.
Correct conclusion in context, following correct work,
level of uncertainty in language.
A0 if hypotheses wrong way round or missing.

6

[Maximum number: 9]

A random sample of 50 values of the continuous random variable X was taken. These values are summarised in the following table.

Table for Question 3 — CAIE A-Level Further Math AS

It is required to test the goodness of fit of the distribution with probability density function f given by

f(x)= \begin{cases}\frac{1}{24}\left(\frac{4}{x^{2}}+x^{2}\right) & 1 \leqslant x \leqslant 4 \\ 0 & \text { otherwise }\end{cases}

The expected frequencies, correct to 4 decimal places, are given in the following table.

(a)

Show that a=4.6007 and find the value of b.

[ 3 ]

$\int_{1.5}^{2} \frac{1}{24}\left(\frac{4}{x^{2}}+x^{2}\right) \mathrm{d} x=\frac{1}{24}\left[-\frac{4}{x}+\frac{x^{3}}{3}\right]$
Integration with correct powers and correct limits
seen.

$a=50 \times 0.092014=4.6007$
AG

$\frac{53}{576}$ or $\frac{1325}{288}$ or 0.092014 or 4.60069 seen

b=11.4211
Or 11.421(0)

3

(b)

Carry out a goodness of fit test, at the 10% significance level, to test whether f is a satisfactory model for the data.

[ 6 ]

Combine first two columns: 6, 9.0278
Must be seen, or implied by 1.0155 .

Calculate value of chi-squared: 1.0155+0.5715+0.7661+0.2183+0.5885
At least 2 correct values (at least 3sf) or
expressions seen.
Allow columns not combined or three columns
combined.

3.16
CWO.

Dependent on M1M1 scored. SCB1 for 3.16 with no working

$\mathrm{H}_{0}: \mathrm{f}$ is a good fit for the data

$\mathrm{H}_{1}$ : f is not a good fit for the data
Tabular value of chi-squared: 7.78

' 3.16 ' < 7.78 and accept $\mathrm{H}_{0}$
Correct tabular value: allow correct FT value if
columns not combined (9.236) or three columns
combined (6.251).

Insufficient evidence to suggest that f is not a good fit for the data

Condone: sufficient evidence to suggest that f is a good fit for the data.
Correct conclusion in context, following correct
work, level of uncertainty in language (not 'prove',
not 'there is no evidence'), no contradictions.
A0 if hypotheses reversed

6

[Maximum number: 7]

(a)

State briefly the circumstances under which a non-parametric test of significance should be used rather than a parametric test.

The level of pollution in a river was measured at 12 randomly chosen locations. The results, in suitable units, are shown below, where higher values represent greater pollution.

Table for Question (a) — CAIE A-Level Further Math AS

[ 1 ]

(b)

Use a Wilcoxon signed-rank test to test whether the average pollution level in the river is more than 6.00. Use a 5% significance level.

[ 6 ]

$\mathrm{H}_{0}$ : population median is $6.00, \mathrm{H}_{1}$ : population median is greater than 6.00

1
Both hypotheses stated

Calculate deviations and resulting signed ranks

1
Devs & -0.38 & -0.27 & 0.55 & 0.81 & 0.10 & -0.25 & -0.13 & 0.47 & -0.14 & 0.26 & 0.99 & -0.09

Rank & -8 & -7 & 10 & 11 & 2 & -5 & -3 & 9 & -4 & 6 & 12 & -1

1
Test statistic T=8+7+5+3+4+1=28

1
Compare with correct critical value 17

1
Conclusion: accept $\mathrm{H}_{0}$ ;
insufficient evidence that the median is greater than
6.00

1

A1FT

Conclusion to be stated in context, not just 'not
significant';
follow through their value of T

6

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Covered syllabus points

Question 1

Question (a)

Question (b)

Question 2

Question 3

Question (a)

Question (b)

Question 4

Question (a)

Question (b)