IB Maths AI SL 4 Statistics and Probability Questions

Practise IB Maths AI SL statistics and probability through data analysis, distributions, regression and technology-supported models with clear interpretation.

Exam points

Collect, present and summarise data while evaluating validity, bias, outliers, reliability and uncertainty.
Interpret regression and correlation, compare models and make predictions without confusing association with causation.
Model probability with conditional events, random variables, expected value and standard distributions.
Use statistical inference and hypothesis tests to evaluate claims, p-values, confidence/uncertainty and limitations.

Covered syllabus points

4.1 Statistics and probability - SL content

[Maximum number: 18]

Elsie, a librarian, wants to investigate the length of time, T minutes, that people spent in her library on a particular day.

(a)

State whether the variable T is discrete or continuous.

[ 1 ]

(b)

Elsie's data for 160 people who visited the library on that particular day is shown in the following table.

Find the value of k.

[ 2 ]

(c)

Write down the modal class.

[ 1 ]

$20 \leq T<40 \quad \boldsymbol{A} 1$

(d)

Write down the mid-interval value for this class.

[ 2 ]

(e)

Use Elsie's data to calculate an estimate of the mean time that people spent in the library.

[ 2 ]

(f)

Using the table, write down the maximum possible number of people who spent 35 minutes or less in the library on that day.

Elsie assumes her data to be representative of future visitors to the library.

[ 1 ]

(g)

Find the probability a visitor spends at least 60 minutes in the library.

[ 2 ]

$\quad \frac{22}{160}\left[0.138,0.1375,13.75 \%, \frac{11}{80}\right]$

Note: Award A1 for correct numerator, A1 for correct denominator.

(h)

The following box and whisker diagram shows the times, in minutes, that the 160 visitors spent in the library.

Figure for Question (h) — IB Maths AI SL

Write down the median time spent in the library.

[ 1 ]

(i)

Find the interquartile range.

[ 2 ]

$\quad 50-16$
Marking guidance:

[1 mark]
[2 marks]

(j)

Hence show that the longest time that a person spent in the library is not an outlier.

Elsie believes the box and whisker diagram indicates that the times spent in the library are not normally distributed.

[ 3 ]

correct substitution into outlier formula
$50+1.5 \times 34$
=101 A1
92<101 OR highest value on diagram <101
not an outlier

Note: Award R1 for their correct comparison. Follow through from their part (h).

Marking guidance:

Award R0 if their conclusion is "it is an outlier", this contradicts Elsie's belief.

(k)

Identify one feature of the box and whisker diagram which might support Elsie's belief.

[ 1 ]

[Maximum number: 5]

The mean annual temperatures for Earth, recorded at fifty-year intervals, are shown in the table.

Tami creates a linear model for this data by finding the equation of the straight line passing through the points with coordinates (1708,8.73) and (1958,9.45).

(a)

Thandizo uses linear regression to obtain a model for the data.

[ 3 ]

(i)

Find the equation of the regression line y on x.

[ 2 ]

$y=0.00256 x+4.46(0.00255714 \ldots x+4.46454 \ldots)$
(M1)A1

Note: Award (M1)A0 for answers that show the correct method, but are presented incorrectly (e.g. no " y= " or truncated values etc.). Accept 4.465 as the correct answer to 4 sf .

(ii)

Find the value of r, the Pearson's product-moment correlation coefficient.

[ 1 ]

(b)

Use Thandizo's model to estimate the mean annual temperature in the year 2000.

[ 2 ]

attempt to substitute 2000 into their part (e)(i)
$0.00255714 \ldots \times 2000+4.46454 \ldots$ $=9.58\left({ }^{\circ} \mathrm{C}\right)\left(9.57882 \ldots\left({ }^{\circ} \mathrm{C}\right)\right)$

Note: Award A1 for 9.57 from $0.00255714 \times 2000+4.46$ .

[Maximum number: 17]

A large international sports tournament tests their athletes for banned substances.
They interpret a positive test result as meaning that the athlete uses banned substances. A negative result means that they do not.
The probability that an athlete uses banned substances is estimated to be 0.06 .
If an athlete uses banned substances, the probability that they will test positive is 0.71 .
If an athlete does not use banned substances, the probability that they will test negative is 0.98 .

(a)

Using the information given, copy (into your answer booklet) and complete the following tree diagram.

Figure for Question (a) — IB Maths AI SL

[ 2 ]

(b)

Determine the probability that a randomly selected athlete does not use banned substances and tests negative.

[ 2 ]

multiplication of two probabilities along the tree diagram
$0.94 \times 0.98$
=0.921(0.9212,92.1 %, 92.12 %)

Note: Do not accept the 2sf value for the final A1.

(c)

If two athletes are selected at random, calculate the probability that both athletes do not use banned substances and both test negative.

[ 2 ]

$(0.9212)^{2}$

=0.849(0.848609 \ldots, 84.9 \%, 84.8609 \ldots \%)

Note: Accept an answer of 0.848 (0.848241) from use of 3 sf answer from part (b)(i).

(d)

Calculate the probability that a randomly selected athlete will receive an incorrect test result.

[ 3 ]

$0.94 \times 0.02+0.06 \times 0.29$
(A1)(M1)

Note: Award A1 for two correct products from their tree diagram seen, M1 for the addition of their two products.

$0.0362\ (3.62\%)$

(e)

A random sample of 1300 athletes at the tournament are selected for testing. Calculate the expected number of athletes in the sample that will receive an incorrect test result.

[ 2 ]

multiplying their part (c)(i) by 1300

$0.0362 \times 1300$

47.1 (47.06)

Note: Accept the 2 sf value of 47 for the final A1.

(f)

Team X are competing in the tournament. There are 20 athletes in this team. It is known that none of the athletes in Team X use banned substances.

Calculate the probability that none of the athletes in Team X will test positive.

[ 4 ]

$\quad p=0.02$ OR p=0.98
recognition of binomial probability with n=20

\mathrm{P}(X=0) \text { OR } \mathrm{P}(X=20)

0.668 (0.667607...)

Note: Award (A1)(M1)(M1)A0 for an answer of 0.667 .
$0.98^{20}=0.668(0.667607 \ldots)$ is awarded full marks.

(g)

Determine the probability that more than 2 athletes in Team X will test positive.

[ 2 ]

$\quad \mathrm{P}(X \geq 3)$ OR $\mathrm{P}(X \leq 17)$
0.00707 (0.00706869...)

Note: Award (M1)A0 for an answer of 0.00706 .
FT from their value of p in part (d)

[Maximum number: 25]

A recent study found that the heights of Dutch women can be modelled by a normal distribution with mean 170.7 cm and standard deviation 6.3 cm .
A Dutch woman is chosen at random.

(a)

Calculate the probability that her height is

[ 8 ]

(i)

less than 160 cm .

[ 4 ]

P(X<160) OR labelled sketch of region O R calc syntax with correct bounds
(M1)
Note: Accept either zero or a large negative value as the lower bound.

=0.0447(0.0447149 \ldots, 4.47 \%)

(ii)

between 160 cm and 170 cm .
27 % of Dutch women have a height of more than h metres.

[ 4 ]

P(160<X<170) OR labelled sketch of region O R calc syntax with correct bounds
(M1)

=0.411 \quad(0.411049 \ldots, 41.1 \%)

Note: Award A0A2 for answers of 0.045 and 0.41 both given to 2 sf.

(b)

Calculate the value of h.

[ 2 ]

$\quad \mathrm{P}(X>h)=0.27$ OR labelled sketch of region OR calc syntax with correct bounds
$=1.75(\mathrm{~m})(1.74560 \ldots)$
Note: Accept 175 (cm).

(c)

Janneke selects a random sample of 200 Dutch women from Amsterdam and measures their heights. She wants to determine whether this sample could have been chosen from a normally distributed population with mean of 170.7 cm and standard deviation of 6.3 cm .

She performs a $\chi^{2}$ goodness of fit test at the 5 % significance level. She begins by creating the following frequency table.

Calculate, correct to four significant figures, the value of

[ 4 ]

(i)

[ 2 ]

(ii)

b.

The hypotheses for Janneke's test are
$\mathrm{H}_{0}$ : the heights are drawn from a normally distributed population with mean 170.7 cm and standard deviation 6.3 cm
$\mathrm{H}_{1}$ : the heights are not drawn from a normally distributed population with mean 170.7 cm and standard deviation 6.3 cm

[ 2 ]

(d)

Write down the degrees of freedom for this test.

The critical value for this test is 7.815 .

[ 1 ]

(e)

Perform the $\chi^{2}$ goodness of fit test and state your conclusion, justifying your reasoning.

Gundega claims that, on average, Latvian women are taller than Dutch women.
Random samples of 10 Latvian women and 10 Dutch women are chosen, and their heights are measured.

Gundega performs a t-test at the 5 % significance level. It is assumed that the populations are normally distributed and have equal variances.

[ 4 ]

p-value $=0.616(0.615583 \ldots)$ OR $\chi^{2}=1.80(1.79702 \ldots)$

Note: Award A1A0 if the p-value or $\chi^{2}$ -value is given correct to 2 dp .
$0.615583 \ldots>0.05$ OR $1.79702 \ldots<7.815$
EITHER
fail to reject the null hypothesis

OR
the heights are normally distributed with mean 170.7 cm and standard deviation 6.3 cm
Note: Do not award R0A1. Condone "accept" in place of "fail to reject".
The R1A1 can be awarded as follow through within part (e) from their (explicitly labelled) p-value or $\chi^{2}$ -value. Accept comparison in words.

(f)

Write down the null and alternative hypotheses for this test.

[ 2 ]

EITHER
$\mathrm{H}_{0}: \mu_{L}=\mu_{D} \quad \boldsymbol{A} \mathbf{1}$ $\mathrm{H}_{1}: \mu_{L}>\mu_{D} \quad$ A1
OR
$\mathrm{H}_{0}$ : The (population) mean height of Latvian women is equal to the (population) mean height of Dutch women A1
$\mathrm{H}_{1}$ : The (population) mean height of Latvian women is greater than the (population) mean height of Dutch women A1

Note: Award at most A0A1 if the hypotheses explicitly refer to the "sample" and not the population. For $\mathrm{H}_{0}: m_{1}=m_{2}$ and $\mathrm{H}_{1}: m_{1}>m_{2}$ award AOA1.

Marking guidance:

(g)

Perform the t-test and state the conclusion, justifying your reasoning.

[ 4 ]

p-value $=0.673(0.673205 \ldots) \quad \boldsymbol{A} \mathbf{2}$

Note: In this question the p-value is the same 3 sf value for unpooled GDC settings so will be awarded A2.
If using a two-tailed test, the answer is p-value $=0.654(0.653589 \ldots)$ ;
award A1
if alternative hypothesis was correct or A2 if it follows through correctly from their alternative hypothesis (i.e. two-tailed test was penalized in part (f)).
$0.673205>0.05 \quad \boldsymbol{R}$
fail to reject the null hypothesis (Gundega is not correct) A1

Note: Do not award R0A1. Condone "accept" in place of "fail to reject".
The R1A1 can be awarded as follow through within part (g) from their (explicitly labelled) p-value. Accept comparison in words.

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Covered syllabus points

Question 1

Question (a)

Question (b)

Question (c)

Question (d)

Question (e)

Question (f)

Question (g)

Question (h)

Question (i)

Question (j)

Question (k)

Question 2

Question (a)

Question (i)

Question (ii)

Question (b)

Question 3

Question (a)

Question (b)

Question (c)

Question (d)

Question (e)

Question (f)

Question (g)

Question 4

Question (a)

Question (i)

Question (ii)

Question (b)

Question (c)

Question (i)

Question (ii)

Question (d)

Question (e)

Question (f)

Question (g)