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: 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: 7]

A town council has published its plans for redeveloping the town centre and residents are being asked whether they approve or disapprove. A random sample of 250 responses has been selected from residents in the four main streets in the town: North, East, South and West Streets. The results are shown in the table.

Table for Question 2 — CAIE A-Level Further Math AS

Test, at the 5% significance level, whether the opinions of the residents are independent of the streets on which they live.

$\mathrm{H}_{0}$ : opinion is independent of street $\mathrm{H}_{1}$ : opinion is not independent of street
33 & 32.24 & 54 & 57.66 & 42 & 43.4 & 26 & 21.7 & 155

19 & 19.76 & 39 & 35.34 & 28 & 26.6 & 9 & 13.3 & 95

52 & & 93 & & 70 & & 35 & & 250
At least 2 correct values or expressions.
6 correct values or expressions.

Calculate chi-squared values:
0.0179+0.2323+0.0452+0.8521+0.0292+0.3790+0.0737+1.3902
Correct to 3 decimal places.
At least 2 correct values or expressions.

3.02
SC B1 3.02 following M1M0
SC B2 3.02 following M0M0

Tabular value, 3 degrees of freedom 7.815: ' 3.02 ' <7.815, accept $\mathrm{H}_{0}$
Allow 'not significant'.

Insufficient evidence to suggest that opinion depends on street.
CWO. Correct conclusion in context, following correct
work, level of uncertainty in language.
A0 if hypotheses wrong way round or missing.

7

[Maximum number: 10]

A school is conducting an experiment to see whether the distance that children can throw a ball increases in hot weather. On a cold day, all the children at the school were asked to throw a ball as far as possible. The distances thrown were measured and recorded. The median distance thrown by a random sample of 25 of the children was 22.0 m . The children were asked to throw the ball again on a hot day. The distances thrown by the same 25 children were measured and recorded and these distances, in m , are shown below.

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

The teacher claims that on average the distances thrown will be further when it is hot.
Carry out a Wilcoxon signed-rank test, at the 5% significance level, to test whether the data supports the teacher's claim.
[10]

$\mathrm{H}_{0}$ : population medians are equal

$\mathrm{H}_{1}$ : population median after > population median before
Do not accept 'differences between population medians < 0 ' unless difference defined.

Allow $\mathrm{H}_{0}: m=22, \mathrm{H}_{1}: m>22.0$

-0.8 & -7 & 1.5 & 12 & 0.9 & 8 & -3.4 & -19 & -2.6 & -18

0.1 & 1 & 4.5 & 21 & -1.8 & -14 & 3.7 & 20 & -1.4 & -11

0.3 & 3 & -4.6 & -22 & 0.2 & 2 & 5 & 23 & 1.9 & 15

6.2 & 25 & 0.6 & 5 & 5.2 & 24 & 1 & 9 & 1.7 & 13

-2.2 & -16 & 0.7 & 6 & 1.3 & 10 & -0.5 & -4 & 2.3 & 17
Signed differences, allow at most 4 errors
Attempt at ranks (ignore signs).

( $\mathrm{W}_{+}=214$ ) $\mathrm{W}_{-}=111$
Cao identified, or used, as test statistic.

Normal: mean $=\frac{1}{4} n(n+1)=\frac{1}{4} \times 25 \times 26[=162.5]$
Variance $=\frac{1}{24} n(n+1)(2 n+1)=\frac{1}{24} \times 25 \times 26 \times 51[=1381.25]$
z - value: $\frac{111.5-162.5}{\sqrt{1381.25}}$
Allow missing or incorrect continuity correction.
Their 111 must come from ranks.

-1.37
CAO

Tabular value is -1.645: ' -1.37 ' >-1.645, or 0.915<0.95 oe, accept $\mathrm{H}_{0}$ ,
Consistent signs. Allow 'not significant'.

Insufficient evidence to support the teacher's claim

Insufficient evidence to suggest that the distances thrown are further when it is hot
All correct. Correct conclusion in context, following correct work, level of uncertainty in language.

A0 if hypotheses wrong way round or missing

10

[Maximum number: 6]

Shane is studying the lengths of the tails of male red kangaroos. He takes a random sample of 14 male red kangaroos and measures the length of the tail, $x \mathrm{~m}$ , for each kangaroo. He then calculates a 90% confidence interval for the population mean tail length, $\mu \mathrm{m}$ , of male red kangaroos. He assumes that the tail lengths are normally distributed and finds that $1.11 \leqslant \mu \leqslant 1.14$ .

Find the values of $\sum x$ and $\sum x^{2}$ for this sample.

$\bar{x}+1.771 \sqrt{\frac{s^{2}}{14}}=1.14$ or $\bar{x}-1.771 \sqrt{\frac{s^{2}}{14}}=1.11$
SOI
Allow incorrect t-value, not z-value.

[Add: $\left.\bar{x}=\frac{1}{2}(1.14+1.11)=1.125\right] \quad \sum x=15.75$
Does not depend on use of a t-value.

Subtract or substitute: $\sqrt{\frac{s^{2}}{14}}=\frac{1}{2}\left(\frac{1.14-1.11}{1.771}\right)$
Allow incorrect t-value, but not a z-value.

$s^{2}=0.00100[4]$ or s=0.0316[9] $\frac{450}{448063}$ , implied by correct final answer.

$\sum x^{2}=13 s^{2}+\frac{\left(\sum x\right)^{2}}{14}$

М1

OE

$\sum x^{2}=17.7$ (3)
CWO

6

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Covered syllabus points

Question 1

Question 2

Question 3

Question 4