IB Math AA SL 4.1 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.

(a)

Find the value of a.

[ 2 ]

summing probabilities and equating to 1

\begin{aligned} & 1.5 a+2 a+0.281+a+0.026=1 \\ & (a=) 0.154 \end{aligned}

(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: 15]

The following cumulative frequency graph shows the monthly income, I dollars, of 2000 families.

(a)

Find the median monthly income.

[ 2 ]

recognizing that the median is at half the total frequency
(M1)
eg $\frac{2000}{2}$
m=2500 (dollars)

(b)

Write down the number of families who have a monthly income of 2000 dollars or less.

[ 1 ]

(c)

Find the number of families who have a monthly income of more than 4000 dollars.

The 2000 families live in two different types of housing. The following table gives information about the number of families living in each type of housing and their monthly income I.

[ 3 ]

(d)

Find the value of p.

[ 2 ]

(e)

A family is chosen at random.

[ 4 ]

(i)

Find the probability that this family lives in an apartment.

[ 2 ]

correct working
eg $\quad 436+765+28$
0.6145 (exact)
$\frac{1229}{2000}, 0.615[0.614,0.615]$

(ii)

Find the probability that this family lives in an apartment, given that its monthly income is greater than 4000 dollars.

[ 2 ]

correct working/probability for number of families
eg $\quad 122+28, \frac{150}{2000}, 0.075$
0.186666
$\frac{28}{150}\left(=\frac{14}{75}\right), 0.187[0.186,0.187]$

(f)

Estimate the mean monthly income for families living in a villa.

[ 3 ]

evidence of using correct mid-interval values (1500,3000,4500)
attempt to substitute into $\frac{\sum f x}{\sum f}$
eg $\frac{1500 \times 64+3000 \times p+4500 \times 122}{64+585+122}$
3112.84

3110 [3110, 3120] (dollars)

[Maximum number: 8]

In a study, measurements for arm span, $A \mathrm{~cm}$ , and foot length, $F \mathrm{~cm}$ , are taken from a large group of adults.
For this group, the regression line of F on A is found to be F=0.335 A-32.6, and the regression line of A on F is found to be A=2.89 F+99.3. Each regression line passes through the mean point.

(a)

By using an appropriate regression line, find an estimate of the arm span for an adult with a foot length of 19.8 cm .

[ 2 ]

attempt to substitute F=19.8 into the regression line for A on F
A=2.89(19.8)+99.3
$=156.522(\mathrm{~cm})$
arm span $=157(\mathrm{~cm})$

Note: Award M0A0 for choosing the wrong regression line to get
$A=156.417 \ldots$ so A=156.

(b)

For this group of adults, find the mean arm span and the mean foot length.

The heights, $H \mathrm{~cm}$ , of adults in the group can be modelled by a normal distribution with mean 163 cm and standard deviation $\sigma \mathrm{cm}$ .

It is found that 88 % of the group have a height between 153 cm and 173 cm .

[ 3 ]

recognition that the lines intersect at the mean point (may be seen on a sketch)
$2.89 F+99.3=\frac{F+32.6}{0.335}$ OR $0.335 A-32.6=\frac{A-99.3}{2.89}$
159.686… or $20.8948 \ldots$
the mean arm span $=160(\mathrm{~cm})$ , the mean foot length $=20.9(\mathrm{~cm})$

(c)

The heights, $H\ \mathrm{cm}$ , of adults in the group can be modelled by a normal distribution with mean 163 cm and standard deviation $\sigma\ \mathrm{cm}$ . It is found that 88 % of the group have a height between 153 cm and 173 cm.

Find the value of $\sigma$ .

[ 3 ]

METHOD 1 recognition of symmetry of interval around mean (may be seen on a sketch)
P(H<153)=0.06 OR P(H<173)=0.94 OR equivalent
$\frac{153-163}{\sigma}=-1.55477 \ldots$ OR $\frac{173-163}{\sigma}=1.55477 \ldots$ $\sigma=6.43181 \ldots$ $\sigma=6.43(\mathrm{~cm})$

METHOD 2
attempt to find $\sigma$ by equating an appropriate correct normal CDF function to 0.88
(or e.g. 0.06 or 0.94 )
$\sigma=6.43181 \ldots$ $\sigma=6.43(\mathrm{~cm})$

Note: Accept use of calculator notation e.g.
normcdf $(153,173,163, \sigma)=0.88$

[Maximum number: 1]

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.

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

Simple random / Systematic / Convenience / Quota / Stratified

[Maximum number: 3]

The following cumulative frequency graph shows the monthly income, I dollars, of 2000 families.

Estimate the mean monthly income for families living in a villa.

evidence of using correct mid-interval values (1500,3000,4500)
attempt to substitute into $\frac{\sum f x}{\sum f}$
eg $\frac{1500 \times 64+3000 \times p+4500 \times 122}{64+585+122}$
3112.84

3110 [3110, 3120] (dollars)

[Maximum number: 3]

In a study, measurements for arm span, $A \mathrm{~cm}$ , and foot length, $F \mathrm{~cm}$ , are taken from a large group of adults.
For this group, the regression line of F on A is found to be F=0.335 A-32.6, and the regression line of A on F is found to be A=2.89 F+99.3. Each regression line passes through the mean point.

For this group of adults, find the mean arm span and the mean foot length.

The heights, $H \mathrm{~cm}$ , of adults in the group can be modelled by a normal distribution with mean 163 cm and standard deviation $\sigma \mathrm{cm}$ .

It is found that 88 % of the group have a height between 153 cm and 173 cm .

recognition that the lines intersect at the mean point (may be seen on a sketch)
$2.89 F+99.3=\frac{F+32.6}{0.335}$ OR $0.335 A-32.6=\frac{A-99.3}{2.89}$
159.686… or $20.8948 \ldots$
the mean arm span $=160(\mathrm{~cm})$ , the mean foot length $=20.9(\mathrm{~cm})$

[Maximum number: 2]

The following cumulative frequency graph shows the monthly income, I dollars, of 2000 families.

A family is chosen at random.

Find the probability that this family lives in an apartment.

correct working
eg $\quad 436+765+28$
0.6145 (exact)
$\frac{1229}{2000}, 0.615[0.614,0.615]$

[Maximum number: 6]

A nationwide study on reaction time is conducted on participants in two age groups. The participants in Group X are less than 40 years old. Their reaction times are normally distributed with mean 0.489 seconds and standard deviation 0.07 seconds.

A randomly selected participant has a reaction time greater than 0.65 seconds. Find the probability that the participant is in Group X.

correct work for P (group X and t>0.65 ) or P (group Y and t>0.65 ) (may be seen anywhere)
eg P( group X $) \times \mathrm{P}(t>0.65 \mid \mathrm{X}), \mathrm{P}(X \cap t>0.65)=0.0107 \times 0.38(=0.004075)$ , $\mathrm{P}(Y \cap t>0.65)=0.396 \times 0.62$
recognizing conditional probability (seen anywhere)
eg $\mathrm{P}(\mathrm{X} \mid t>0.65), \mathrm{P}(A \mid B)=\frac{\mathrm{P}(A \cap B)}{\mathrm{P}(B)}$
valid approach to find P(t>0.65)

Figure for Answer to Question 8 — IB Maths AA SL

correct work for P(t>0.65)
eg $\quad 0.0107 \times 0.38+0.396 \times 0.62,0.249595$
correct substitution into conditional probability formula
eg $\frac{0.0107 \times 0.38}{0.0107 \times 0.38+0.396 \times 0.62}, \frac{0.004075}{0.249595}$
0.016327
$\mathrm{P}(\mathrm{X} \mid t>0.65)=0.0163270$

[Maximum number: 3]

A nationwide study on reaction time is conducted on participants in two age groups. The participants in Group X are less than 40 years old. Their reaction times are normally distributed with mean 0.489 seconds and standard deviation 0.07 seconds.

Ten of the participants with reaction times greater than 0.65 are selected at random. Find the probability that at least two of them are in Group X.

recognizing binomial probability
eg $\quad X \sim \mathrm{~B}(n, p),\binom{n}{r} p^{r} q^{n-r},(0.016327)^{2}(0.983672)^{8},\binom{10}{2}$
valid approach
eg $\mathrm{P}(X \geq 2)=1-\mathrm{P}(X \leq 1), 1-\mathrm{P}(X<a)$ , summing terms from 2 to 10 (accept binomcdf( 10,0.0163,2,10 ) )
0.010994
$\mathrm{P}(X \geq 2)=0.0110$

[Maximum number: 2]

A nationwide study on reaction time is conducted on participants in two age groups. The participants in Group X are less than 40 years old. Their reaction times are normally distributed with mean 0.489 seconds and standard deviation 0.07 seconds.

A person is selected at random from Group X . Find the probability that their reaction time is greater than 0.65 seconds.

The participants in Group Y are 40 years or older. Their reaction times are normally distributed with mean 0.592 seconds and standard deviation $\sigma$ seconds.

[Maximum number: 2]

In a study, measurements for arm span, $A \mathrm{~cm}$ , and foot length, $F \mathrm{~cm}$ , are taken from a large group of adults.
For this group, the regression line of F on A is found to be F=0.335 A-32.6, and the regression line of A on F is found to be A=2.89 F+99.3. Each regression line passes through the mean point.

By using an appropriate regression line, find an estimate of the arm span for an adult with a foot length of 19.8 cm .

attempt to substitute F=19.8 into the regression line for A on F
A=2.89(19.8)+99.3
$=156.522(\mathrm{~cm})$
arm span $=157(\mathrm{~cm})$

Note: Award M0A0 for choosing the wrong regression line to get
$A=156.417 \ldots$ so A=156.

[Maximum number: 2]

The following cumulative frequency graph shows the monthly income, I dollars, of 2000 families.

A family is chosen at random.

Find the probability that this family lives in an apartment, given that its monthly income is greater than 4000 dollars.

correct working/probability for number of families
eg $\quad 122+28, \frac{150}{2000}, 0.075$
0.186666
$\frac{28}{150}\left(=\frac{14}{75}\right), 0.187[0.186,0.187]$

[Maximum number: 4]

A nationwide study on reaction time is conducted on participants in two age groups. The participants in Group X are less than 40 years old. Their reaction times are normally distributed with mean 0.489 seconds and standard deviation 0.07 seconds.

The probability that the reaction time of a person in Group Y is greater than 0.65 seconds is 0.396 . Find the value of $\sigma$ .

In the study, 38 % of the participants are in Group X.

correct z-value
0.263714...
evidence of appropriate approach
eg $\quad \frac{0.65-0.592}{\sigma}, 0.264=\frac{x-\mu}{\sigma}$
correct substitution
eg $\quad 0.263714=\frac{0.65-0.592}{\sigma}, \sigma=\frac{0.65-0.592}{0.264}$
0.219934
$\sigma=0.220$

IB Maths AA SL 4.1 Statistics and Probability Sl Content Questions

Exam points

Covered syllabus points

Covered syllabus points

Question 1

Question (a)

Question (b)

Question (c)

Question (d)

Question 2

Question (a)

Question (b)

Question (c)

Question (d)

Question (e)

Question (i)

Question (ii)

Question (f)

Question 3

Question (a)

Question (b)

Question (c)

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Question 11

Question 12

Question 13