IB Maths AA HL 4 Statistics & Probability

[Maximum number: 6]

A supermarket analyses the shopping habits of its customers.
The number of times, X, each customer visits the supermarket in a week is given by the following probability distribution.

Table for Question 4 Statistics and probability question 1 — IB Maths AA HL

(a)

Find the value of a.

[ 2 ]

(b)

Write down the mode of X.

[ 1 ]

(c)

Find the mean of X.

The manager wants to know why customers come to their supermarket. They survey the first 50 customers to arrive at the supermarket on a particular day.

[ 2 ]

(d)

Identify which one of the following best describes the manager's sampling method. Circle your answer.

Simple random / Systematic / Convenience / Quota / Stratified

[ 1 ]

[Maximum number: 18]

Amanda enters data from surveys into a database. It can be assumed that the accuracy of any survey entered is independent of all other surveys entered.
From previous records, it is known that Amanda enters 8 % of the surveys inaccurately.

(a)

On a particular day Amanda enters data from 50 surveys.

[ 5 ]

(i)

Find the probability that Amanda entered at most six surveys inaccurately.

[ 2 ]

recognizing binomial distribution
$E \sim \mathrm{~B}(50,0.08)$ , where E is the number of inaccurate surveys
$\mathrm{P}(E \leq 6)=0.898128 \ldots$ $(\mathrm{P}(E \leq 6)=) 0.898$ (accept 89.8 % )

(ii)

Given that at most six surveys were entered inaccurately, find the probability that exactly four surveys were entered inaccurately.

On a different day Amanda enters data from n surveys. On this day, the probability that at most six surveys were entered inaccurately is approximately 0.367 .

[ 3 ]

recognition of conditional probability

Note: Recognition must be shown in context either in words or symbols, not just $\mathrm{P}(A \mid B)$ .

\begin{aligned} & \mathrm{P}(E=4 \mid E \leq 6)=\frac{\mathrm{P}(E=4)}{\mathrm{P}(E \leq 6)} \\ & =\frac{0.203654 \ldots}{0.898128 \ldots} \\ & =0.227 \text { (accept } 22.7 \% \text { ) } \end{aligned}

(b)

Find the value of n.

Bryce and Carmen also enter data from surveys into the same database. It is known that surveys entered by Bryce and Carmen are inaccurate 6 % and 11 % of the time respectively. It can again be assumed that the accuracy of any survey entered is independent of all other surveys entered.

From the surveys assigned to the three of them, Amanda enters 55 %, Bryce 25 % and Carmen 20\%.

[ 3 ]

use of tables

Note: Award (M1) for at least one correct value of $\mathrm{P}(E \leq 6)$ for a value of n, where $n \neq 50$ .

(n=94 \Rightarrow) \mathrm{P}(E \leq 6)=0.366746 \ldots

n=94

(c)

Find the probability that a randomly selected survey was

[ 6 ]

(i)

entered inaccurately;

[ 3 ]

$\mathrm{P}(I)=\mathrm{P}(I \cap A)+\mathrm{P}(I \cap B)+\mathrm{P}(I \cap C)$ $=0.55 \times 0.08+0.25 \times 0.06+0.2 \times 0.11$
=0.081 (accept 8.1 % )

(ii)

entered by Amanda, given that the survey was entered inaccurately.

The following year, the accuracy of Amanda's and Bryce's work remained the same, as did the percentage of surveys entered by each of the three employees. However, Carmen's accuracy had improved and the probability that she entered a survey inaccurately was now x %.

The probability that a randomly selected survey had been entered inaccurately was now the same as the probability that Carmen made an error when entering a survey.

[ 3 ]

$\mathrm{P}(A \mid I)=\frac{\mathrm{P}(A \cap I)}{\mathrm{P}(I)}$ (or use of Bayes' formula)

\mathrm{P}(A \mid I)=\frac{0.55 \times 0.08}{0.081}

=0.543 \text { (accept } 54.3 \% \text { ) }

(d)

Find the value of x.

[ 4 ]

METHOD 1

\begin{aligned} & \mathrm{P}(I \mid C)=\frac{x}{100} \text { (seen anywhere) } \\ & \mathrm{P}(I)=0.55 \times 0.08+0.25 \times 0.06+0.2 \times \frac{x}{100} \end{aligned}

equating their expression for P(I) to $\frac{x}{100}$

\begin{aligned} & \frac{x}{100}=0.55 \times 0.08+0.25 \times 0.06+0.2 \times \frac{x}{100} \\ & x=7.375 \\ & x=7.38 \text { (accept } 7.38 \% \text { ) } \end{aligned}

METHOD 2

\mathrm{P}(I)=0.55 \times 0.08+0.25 \times 0.06+0.2 \times y

equating their expression for P(I) to y
$y=0.55 \times 0.08+0.25 \times 0.06+0.2 \times y$
y=0.07375
recognizing to multiply their y value by 100
x=7.38 (accept 7.38 % )

[Maximum number: 15]

The continuous random variable X has a probability density function given by

f(x)=\left\{\begin{array}{cc} k \sin \left(\frac{\pi x}{6}\right), & 0 \leq x \leq 6 \\ 0, & \text { otherwise } \end{array}\right.

(a)

Find the value of k.

[ 4 ]

attempt to equate integral to 1 (may appear later)
$k \int_{0}^{6} \sin \left(\frac{\pi x}{6}\right) \mathrm{d} x=1$
correct integral
$k\left[-\frac{6}{\pi} \cos \left(\frac{\pi x}{6}\right)\right]_{0}^{6}=1$
substituting limits
$-\frac{6}{\pi}(-1-1)=\frac{1}{k}$ $k=\frac{\pi}{12}$

(b)

By considering the graph of f write down

[ 3 ]

(i)

the mean of X;

[ 1 ]

(ii)

the median of X;

[ 1 ]

(iii)

the mode of X.

[ 1 ]

(c)

Show that $P(0 \leq X \leq 2)=\frac{1}{4}$ .

[ 4 ]

$\frac{\pi}{12} \int_{0}^{2} \sin \left(\frac{\pi x}{6}\right) \mathrm{d} x$

=\frac{\pi}{12}\left[-\frac{6}{\pi} \cos \left(\frac{\pi x}{6}\right)\right]_{0}^{2}

Note: Accept without the $\frac{\pi}{12}$ at this stage if it is added later.

\begin{aligned} & \frac{\pi}{12}\left[-\frac{6}{\pi}\left(\cos \frac{\pi}{3}-1\right)\right] \\ & =\frac{1}{4} \end{aligned}

(d)

Hence state the interquartile range of X.

[ 2 ]

from (c)(i) $Q_{1}=2$
as the graph is symmetrical about the middle value $x=3 \Rightarrow Q_{3}=4$
so interquartile range is
4-2
=2

(e)

Calculate $P(X \leq 4 \mid X \geq 3)$ .

[ 2 ]

$\quad P(X \leq 4 \mid X \geq 3)=\frac{P(3 \leq X \leq 4)}{P(X \geq 3)}$ $=\frac{\frac{1}{4}}{\frac{1}{2}}$
(M1)
$=\frac{1}{2}$
A1

[Maximum number: 14]

The random variable X has probability distribution $\operatorname{Po}(8)$ .

(a)

Find P(X=6).

[ 2 ]

(b)

Find $\mathrm{P}(X=6 \mid 5 \leq X \leq 8)$ .

[ 3 ]

$\mathrm{P}(X=6 \mid 5 \leq X \leq 8)=\frac{\mathrm{P}(X=6)}{\mathrm{P}(5 \leq X \leq 8)}=\frac{0.122 \ldots}{0.592 \ldots-0.0996 \ldots} =0.248$
(M1)(A1)
A1

(c)

$\bar{X}$ denotes the sample mean of n>1 independent observations from X.

[ 3 ]

(i)

Write down $\mathrm{E}(\bar{X})$ and $\operatorname{Var}(\bar{X})$ .

[ 2 ]

$\quad \mathrm{E}(\bar{X})=8$

A1
$\operatorname{Var}(\bar{X})=\frac{8}{n}$

(ii)

Hence, give a reason why $\bar{X}$ is not a Poisson distribution.

[ 1 ]

$\quad \mathrm{E}(\bar{X}) \neq \operatorname{Var}(\bar{X})($ for n>1)

R1

Note: Only award the R1 if the two expressions in (b)(i) are different.

(d)

A random sample of 40 observations is taken from the distribution for X.

[ 6 ]

(i)

Find $\mathrm{P}(7.1<\bar{X}<8.5)$ .

[ 3 ]

EITHER

Note: M1 for normality, A1 for parameters.

OR
The expression is equivalent to $P\left(283 \leq \sum X \leq 339\right)$ where $\sum X$ is $\operatorname{Po}(320)$
= 0.840
Note: Accept 284, 340 instead of 283, 339

Accept any answer that rounds correctly to 0.84 or 0.85 .

(ii)

Given that $\mathrm{P}(|\bar{X}-8| \leq k)=0.95$ , find the value of k.

[ 3 ]

EITHER
$k=1.96 \frac{\sigma}{\sqrt{n}}$ or $1.96 \operatorname{std}(\bar{X})$
(M1)(A1)
k=0.877 or $1.96 \sqrt{0.2}$

OR
The expression is equivalent to
$P\left(320-40 k \leq \sum X \leq 320+40 k\right)=0.95$
(M1)
k=0.875

Note: Accept any answer that rounds to 0.87 or 0.88 .
Award M1A0 if modulus sign ignored and answer obtained rounds to 0.74 or 0.75
Total

IB Maths AA HL 4 Statistics and Probability

Covered syllabus points

Covered syllabus points

4 Statistics and probability question 1

Question (a)

Question (b)

Question (c)

Question (d)

4 Statistics and probability question 2

Question (a)

Question (i)

Question (ii)

Question (b)

Question (c)

Question (i)

Question (ii)

Question (d)

4 Statistics and probability question 3

Question (a)

Question (b)

Question (i)

Question (ii)

Question (iii)

Question (c)

Question (d)

Question (e)

4 Statistics and probability question 4

Question (a)

Question (b)

Question (c)

Question (i)

Question (ii)

Question (d)

Question (i)

Question (ii)