# IB Math Analysis & Approaches Questionbank-Topic: SL 4.1 Concepts of population, sample, random sample, discrete and continuous data SL Paper 1

## Question

The following box-and-whisker plot shows the number of text messages sent by students in a school on a particular day.

Find the value of the interquartile range.

[2]
a.

One student sent k text messages, where k > 11 . Given that k is an outlier, find the least value of k.

[4]
b.

## Markscheme

recognizing Q1 or Q3 (seen anywhere)     (M1)

eg    4,11 , indicated on diagram

IQR = 7     A1 N2

[2 marks]

a.

recognizing the need to find 1.5 IQR     (M1)

eg   1.5 × IQR, 1.5 × 7

valid approach to find    (M1)

eg   10.5 + 11, 1.5 × IQR + Q3

21.5     (A1)

k = 22     A1 N3

Note: If no working shown, award N2 for an answer of 21.5.

[4 marks]

b.

MAA SL 4.1-4.3 STATISTICS – BASIC CONCEPTS [concise]-lala

### Question

[Maximum mark: 5] [without GDC]
A set of data is 6, 2, 3, 8, 6, 5, 7, 6, 2. Find
(a) the measures of central tendency: mode, median, mean.
(b) the measures of spread: range, interquartile range.

Ans.

mode = 6, median = 6, mean = 5,
range = 6,  IQR = 6.5 – 2.5 = 4

### Question

[Maximum mark: 5] [with GDC]
For the data in Exercise 1,
(a) confirm the results of questions 1(a) and 1(b) by your GDC;
(b) find the measures of spread: standard deviation and variance.

Ans.

standard deviation = 2.05           variance = 2.052

### Question

[Maximum mark: 7] [without GDC]
For the following 200 entries,

find
(a) the measures of central tendency: mode, median, mean.
(b) the measures of spread: range, interquartile range.
(c) the 65th and 70th percentiles

Ans.

mode = 6, median = 6, mean = 6.76
range = 4, IQR = 8.5 – 5 = 3.5
65th percentile = 7.5          70th percentile = 8

### Question

[Maximum mark: 5] [with GDC]
For the data in Exercise 3,
(a) confirm the results of questions 3(a) and 3(b) by your GDC;
(b) find the measures of spread: standard deviation and variance.

Ans.

standard deviation = 1.556                     variance = 1.5562

### Question

[Maximum mark: 7] [without GDC]
For the data presented in the stem and leaf diagram below

(a) write down the first four entries (i.e. the smallest ones)
(b) find the measures of central tendency: mode and median,
(c) find the measures of spread: range and interquartile range

Ans.

(a) 51, 53, 57, 60
(b) mode = 84, median = 78
(c) range = 48, IQR = 84 – 60 = 24

### Question

[Maximum mark: 5] [without GDC]
The following results give the heights of sunflowers in centimetres.
180 184 195 177 175 173 169 167 197 173 166 183 161 195 177
192 161 165
Represent the data by a stem and leaf diagram.

Ans.

### Question

[Maximum mark: 12] [without GDC]
Consider the data 10, 20, 30, 40 with
mean = 25        standard deviation = $$5\sqrt{5}$$        variance = 125
Find the new mean, standard deviation and variance in the following cases

Ans.

mean = 25        standard deviation = $$5\sqrt{5}$$         variance = 125

### Question

[Maximum mark: 7] [with GDC]
Consider the following data

(a) Find the mean by using the appropriate formula.
(b) Write down the modal group
(c) Find the standard deviation and the variance

Ans.

(a) mean = $$\frac{3×16+8×28+13×26+18×14}{84}=10.3$$
(b) modal group: 6-10
(c) standard deviation = 4.90 variance = 24.0

### Question

[Maximum mark: 6] [without GDC]
Three positive integers $$a$$, $$b$$, and $$c$$, where $$a < b < c$$, are such that their median is 11,
their mean is 9 and their range is 10. Find the value of $$a$$.

Ans.

Median = middle value => $$b$$ = 11
Mean = $$\frac{a+b+c}{3}=\frac{a+11+c}{3}=9=>a+11+c=27=>a+c=16$$
Range = $$c – a$$ = 10
Solving equations simultaneously gives $$a$$ = 3

### Question

[Maximum mark: 6] [without GDC]
Let $$a, b, c$$ and $$d$$ be integers such that $$a < b, b < c$$ and $$c = d$$.
The mode of these four numbers is 11. The range is 8. The mean is 8.
Calculate the value of each of the integers $$a, b, c, d$$.

Ans.

$$d$$ = 11;      $$c$$ = 11
$$d – a$$ = 8 ⇒ $$a$$ = 3
$$\frac{3+b+11+11}{4}=8\Rightarrow b=7$$

### Question

[Maximum mark: 6] [without GDC]
Consider the four numbers $$a, b, c, d$$ with $$a ≤ b ≤ c ≤ d$$, where $$a,b,c,d \in \mathbb{Z}$$.
The mean of the four numbers is 4. The mode is 3. The median is 3. The range is 6.
Find the value of $$a$$, of $$b$$, of $$c$$ and of $$d$$.

Ans.

$$b$$ = 3,  $$c$$ = 3
using mean $$\left ( \frac{a+b+c+d}{4}=4 \right )$$ and range ($$d – a$$ = 6)
$$a$$ = 2,    $$d$$ = 8

### Question

[Maximum mark: 6] [with GDC]
The population below is listed in ascending order.
5, 6, 7, 7, 9, 9, $$r$$, 10, s, 13, 13, $$t$$
The median of the population is 9.5. The upper quartile $$Q$$3 is 13.
(a) Write down the value of    (i)     $$r$$                 (ii)       $$s$$
(b) The mean of the population is 10. Find the value of $$t$$.

Ans.

(a) (i) $$r$$ = 10
(ii) $$s$$ = 13
(b) Using $$\frac{\sum x}{12}=10 \Rightarrow t=18$$

### Question

[Maximum mark: 6] [without GDC]
A set of data is
18, 18, 19, 19, 20, 22, 22, 23, 27, 28, 28, 31, 34, 34, 36.
The box and whisker plot for this data is shown below.

(a) Write down the values of A, B, C, D, E
(b) Find the interquartile range.

Ans.

(a) A = 18,    B = 19,     C = 23,     D = 31,    E = 36
(b) IQR = 12

### Question

[Maximum mark: 6] [without GDC]
The box and whisker diagram shown below represents the marks received by 32
students.

(a) Write down the value of the median mark.
(b) Write down the value of the upper quartile.
(c) Estimate the number of students who received a mark greater than 6.

Ans.

(a) 3
(b) 6
(c) 25% scored greater than 6,
0.25 x 32 = 8

### Question

[Maximum mark: 5] [without GDC]
The following diagram is a box and whisker plot for a set of data.

The interquartile range is 20 and the range is 40.
(a) Write down the median value.
(b) Find the value of    (i) $$a$$;          (ii) $$b$$.

Ans.

(a) 18
(b) (i)   10
(ii)   44

### Question

[Maximum mark: 5] [without GDC]
A scientist has 100 female fish and 100 male fish. She measures their lengths to the
nearest cm. These are shown in the following box and whisker diagrams.

(a) Find the range of the lengths of all 200 fish.
(b) Four cumulative frequency graphs are shown below.

Ans.

(a) correct end points
max = 27, min = 4
range = 23
(b) Graph 3

### Question

[Maximum mark: 6] [with GDC]
The 45 students in a class each recorded the number of whole minutes, $$x$$, spent doing
experiments on Monday. The results are $$Σx = 2230$$.
(a) Find the mean number of min the students spent doing experiments on Monday.
Two new students joined the class and reported that they spent 37 minutes and 30
minutes respectively.
(b) Calculate the new mean including these two students.

Ans.

(a) mean = $$\sum \frac{x}{n}\left ( =\frac{2230}{45} \right )\bar{x}=49.6 (Accept\; 50)$$
(b) $$\bar{y}=\frac{\sum y}{n+2}(may \;be\; implied)$$
$$\sum y=2230+37+30$$
$$\bar{y}=\frac{2297}{47}=48.9 (Accept\;49)$$

### Question

[Maximum mark: 4] [without GDC]
The mean of the population x1, x2, … , x25 is $$m$$

(a) Given that $$\sum_{i=1}^{25}x_{i}=300$$ find the value of $$m$$ .

(b) Given that $$m$$ = 10 find the value of  $$\sum_{i=1}^{25}x_{i}$$.

Ans.

(a) m = $$\frac{300}{25}=12$$

(b) $$\sum_{i=1}^{25}x_{i}=25×10=250$$;

### Question

[Maximum mark: 4] [with GDC]
At a conference of 100 mathematicians there are 72 men and 28 women. The men
have a mean height of 1.79 m and the women have a mean height of 1.62 m. Find the
mean height of the 100 mathematicians.

Ans.

Mean = $$\frac{(72×1.79)+(28×1.62)}{100}=1.7424(=1.74\;to\;3\;sf)$$

### Question

[Maximum mark: 6] [with GDC]
From January to September, the mean number of car accidents per month was 630.
From October to December, the mean was 810 accidents per month.
What was the mean number of car accidents per month for the whole year?

Ans.

Jan–Sept     ∑ = 630 × 9 = 5670
Oct–Dec      ∑ = 810 × 3 = 2430

mean = $$\bar{x}=\frac{5670+2430}{12}=675$$

### Question

[Maximum mark: 7] [with / without GDC]
A box contains 100 cards. Each card has a number between one and six written on it.
The following table shows the frequencies for each number.

(a) Calculate the value of $$k$$.
(b) Find     (i) the median;              (ii) the interquartile range.

Ans.

(a) $$\sum f_{i}$$ = 100 ⇒ k = 4
(b) (i) median position 50th item, 26 + 10 + 20 = 56
median = 3
(ii) $$Q$$1 = and $$Q$$3 = 5
interquartile range = 5-1= 4

### Question

[Maximum mark: 4] [with / without GDC]
Given the following frequency distribution,

find
(a) the median;
(b) the mean.

Ans.

(a) middle number of 75 = 38th number
Median = 4
(b) Mean =$$\frac{5+18+48+72+100+42}{75}=\frac{285}{75}=3.8$$

### Question

[Maximum mark: 6] [with GDC]
A standard die is rolled 36 times. The results are shown in the following table.

(a) Write down the standard deviation.
(b) Write down the median score.
(c) Find the interquartile range.

Ans.

(a) σ = 1.61
(b) median = 4.5
(c) $$Q$$1 = 3, $$Q$$3 = 5 (may be seen in a box plot)
IQR = 2

### Question

[Maximum mark: 6] [with / without GDC]
The number of hours of sleep of 21 students are shown in the frequency table below.

Find
(a) the median;
(b) the lower quartile;
(c) the interquartile range.

Ans.

(a) m = 6
(b) $$Q$$1 = 5
(c) $$Q$$3 = 8
IQR = 8 – 5= 3 (accept 5 – 8 or [5, 8])

### Question

[Maximum mark: 6] [without GDC]
The table below shows the marks gained in a test by a group of students.

The median is 3 and the mode is 2. Find the two possible values of $$p$$.

Ans.

List of frequencies with $$p$$ in the middle, $$eg$$ 5 + 10, $$p$$, 6 + 2
$$p$$ > 7, $$p$$ < 10 because 2 is the mode
Possible values of $$p$$ are 8 and 9

### Question

[Maximum mark: 7] [with GDC]
The following table gives the examination grades for 120 students.

(a) Find the value of
(i) $$p$$;                (ii) $$q$$.
(c) Write down the standard deviation.

Ans.

(a) (i) 9 + 25 + 35  OR  34 + 35
$$p$$ = 69
(ii) 109 – 69  OR  120 – (9 + 25 + 35 + 11)
$$q$$ = 40
(b) mean = 3.16
(c) 1.09

### Question

[Maximum mark: 7] [with GDC]
In a school with 125 girls, each student is tested to see how many sit-up exercises (situps)
she can do in one minute. The results are given in the table below.

(a)   (i) Write down the value of $$p$$                  (ii) Find the value of $$q$$.
(b) Find the median number of sit-ups.
(c) Find the mean number of sit-ups.

Ans.

(a) (i)    $$p$$ = 65
(ii) using sum = 125 (or 99 – $$p$$) ⇒ $$q$$ = 34
(b) median position $$e.g.$$ 63rd student, median is 17 (sit-ups)
(c) mean = $$\frac{15(11)+16(21)+17(33)+18(34)+19(18)+20(8)}{125}=\frac{2176}{125}=17.4$$

### Question

[Maximum mark: 6] [with GDC]
The following table shows the mathematics marks scored by students.

The mean mark is 4.6.
(a) Find the value of $$k$$.
(b) Write down the mode.

Ans.

(a) 4.6 = $$\frac{144+4k}{k+30}$$
4.6$$k$$ + 138 = 144 + 4$$k$$
0.6$$k$$ = 6
$$k$$ = 10
(b) Mode = 4

### Question

[Maximum mark: 4] [with GDC]
The table shows the scores of competitors in a competition.

The mean score is 34. Find the value of $$k$$.

Ans.

$$\frac{(10×1)+(20×2)+(30×5)+(40xk)+(50×3)}{k+11}=34$$
$$\frac{40k+350}{k+11}=34\Rightarrow k=4$$

### Question

[Maximum mark: 6] [without GDC]
The following stem and leaf diagram gives the heights in cm of 39 schoolchildren.

(a) State for the height
(i) the lower quartile,              (ii) the median                (iii) the upper quartile.

(b) Draw a box-and-whisker plot of the data using the axis below.

Ans.

(a)    (i) 145               (ii) 157               (iii) 167

(b)

### Question

[Maximum mark: 6] [with GDC]
The histogram below represents the ages of 270 people in a village.

(a) Use the histogram to complete the table below.

(b) Hence, calculate an estimate of the mean age.

Ans.

(a)

(b) Mean = $$\frac{11900}{270}=44.1$$

### Question

[Maximum mark: 6] [without GDC]
A student measured the diameters of 80 snail shells. His results are shown in the
following cumulative frequency graph. The lower quartile (LQ) is 14 mm and is marked
clearly on the graph.

(a) On the graph, mark clearly in the same way and write down the value of
(i) the median;                         (ii) the upper quartile.
(b) Write down the interquartile range.

Ans.

(a) (i) median = 20 (correct lines drawn on graph)
(ii) $$Q$$3 = 24 (correct lines drawn on graph)
(b) IQR = $$Q$$3 – $$Q$$1 = 10 (accept 14 to 24)

### Question

[Maximum mark: 6] [without GDC]
The cumulative frequency curve below shows the heights of 120 basketball players in
centimetres.

Use the curve to estimate
(a) the median height;
(b) the interquartile range.

Ans.

(a) median is 183 (line(s) on graph)
(b) Lower quartile $$Q$$1 = 175 Upper quartile $$Q$$3 = 189
IQR is 14

### Question

[Maximum mark: 6] [without GDC]
A test marked out of 100 is written by 800 students. The cumulative frequency graph
for the marks is given below.

(a) Write down the number of students who scored 40 marks or less on the test.
(b) The middle 50 % of test results lie between marks $$a$$ and $$b$$, $$(a < b)$$. Find a and b.

Ans.

(a) 100 students score 40 marks or fewer. (Lines on graph)
(b) $$a$$ = 55, $$b$$ = 75 (Identifying 200 and 600 , Lines on graph)

### Question

[Maximum mark: 6] [without GDC]
The following is a cumulative frequency diagram for the time t, in minutes, taken by 80

(a) Write down the median.
(b) Find the interquartile range.
(c) Complete the frequency table next to the diagram.

Ans.

(a) median $$m$$ = 32
(b) lower quartile $$Q$$1 = 22, upper quartile $$Q$$3 = 40 ⇒ interquartile range = 18
(c)

### Question

[Maximum mark: 6] [with GDC]
The cumulative frequency graph below shows the heights of 120 girls in a school.

(a) Find
(i) the median; (ii) the interquartile range.
(b) Given that 60% of the girls are taller than $$a$$ cm, find the value of $$a$$.

Ans.

(a) (i) $$m$$ = 165
(ii) $$Q$$1 = 160     $$Q$$2 = 170   ⇒     IQR = 10
(b) 40th percentile, or 48th student (a horizontal line through (0, 48) )
$$a$$ = 163

### Question

[Maximum mark: 6] [without GDC]
The cumulative frequency curve below shows the marks obtained in an examination by
a group of 200 students.

(a) Use the cumulative frequency curve to complete the frequency table below.

(b) Forty percent of the students fail. Find the pass mark.

Ans.

(a)

(b) 40th Percentile ⇒ 80th student fails, (mark 42%)
Pass mark 43% (Accept mark > 42.)

### Question

[Maximum mark: 6] [with GDC]
In the research department of a university, 300 mice were timed as they each ran
through a maze. The results are shown in the cumulative frequency diagram below.

(a) How many mice complete the maze in less than 10 seconds?
(b) Estimate the median time.
(c) Another way of showing the results is the frequency table next to the graph..
(i) Find the value of $$p$$ and the value of $$q$$.
(ii) Calculate an estimate of the mean time.

Ans.

(a) 76 (mice)
(b) 11.2 (seconds)
(c) (i) $$p$$ = 76 – (16 + 22) = 38     $$q$$ = 132 – 76 = 56
(ii) $$x=\frac{7.5×16+….14.5×23}{16+…23}\left ( =\frac{3363}{300} \right )=11.21(=11.2)$$

### Question

[Maximum mark: 6] [without GDC]
The four populations A, B, C and D are the same size and have the same range.
Frequency histograms for the four populations are given below.

(a) Each of the three box and whisker plots below corresponds to one of the four
populations. Write the letter of the correct population under each plot.

(b) Each of the three cumulative frequency diagrams below corresponds to one of
the four populations. Write the letter of the correct population under each
diagram.

Ans.

(a)    D     B      C
(b)    B      A     C

### Question

[Maximum mark: 10] [with GDC]
The following diagram represents the lengths, in cm, of 80 plants grown in a laboratory.

(a) How many plants have lengths in cm between
(i) 50 and 60?                    (ii) 70 and 90?
(b) Calculate estimates for the mean and the st.deviation of the lengths of the plants.
(c) Explain what feature suggests that the median is different than the mean.
(d) The following is an extract from the cumulative frequency table.

Find an estimate for the median of the lengths of the plants.

Ans.

(a)  (i)    10
(ii)    14 + 10 = 24

(b)

(i) µ = 63
(ii) σ = 20.5 (3 sf)

(c) Assymetric diagram/distribution

(d)

### Question

[Maximum mark: 14] [with GDC]
The following is the cumulative frequency curve for the time, t minutes, spent by 150
people in a store on a particular day.

(a) (i) How many people spent less than 5 minutes in the store?
(ii) Find the number of people who spent between 5 and 7 minutes in the store.
(iii) Find the median time spent in the store.
(b) Given that 40% of the people spent longer than $$k$$ minutes, find the value of $$k$$.
(c) (i) Complete the following frequency table.

(ii) Hence, calculate an estimate for the mean time spent in the store.

Ans.

(a)  (i) 50 (accept 49, “fewer than 50”)
(ii) Cumulative frequency (7) = 90, 90 – 50 = 40
(iii) 75th or 75.5th person, median = 6.25 (min), 6 min 15 secs
(b) finding 40% (or 60%) of 150
Number spending less than $$k$$ minutes is (150 – 60) = 90 ⇒ k = 7
(c) (i)

(ii) using mid-interval values (1, 3, 5, 7, 9, 11)
mean = $$\left ( \frac{1×10+3×23+5×37+7×38+9×27+11×15}{150} \right )$$ = 6.25 (min), 6 min 15 secs

### Question

[Maximum mark: 11] [with GDC]
The speeds in km h–1 of cars passing a point on a highway are recorded in the first
table below.

(a) Calculate an estimate of the mean speed of the cars.
(b) The second table gives some of the cumulative frequencies.
(i) Write down the values of $$a$$ and $$b$$.
(ii) On graph paper, construct a cumulative frequency curve to represent this
information. Use a scale of 1 cm for 10 km h–1 on the horizontal axis and a
scale of 1 cm for 20 cars on the vertical axis.
(c) Use your graph to determine
(i) the percentage of cars travelling at a speed in excess of 105 km h–1;
(ii) the speed which is exceeded by 15% of the cars.

Ans.

(a) (Using mid-intervals) $$\bar{v}=\frac{65(7)+75(25)+…+135(5)}{7+25+…+5}=\frac{29450}{300}=98.2 \;km\; h^{-1}$$

(b) (i)   $$a$$ = 165,    $$b$$ = 275

(ii)

(c) (i) Vertical line on graph at 105 km h–1, $$\frac{300-200}{300}$$ × 100% = 33.3(±1.3%)

(ii) 15% of 300 = 45      300 – 45 = 255
Horizontal line on graph at 255 cars
Speed = 114(± 2 km h–1)

### Question

[Maximum mark: 10] [without GDC]
A survey is carried out to find the waiting times for 100 customers at a supermarket.

(a) Calculate an estimate for the mean of the waiting times, by using an appropriate
approximation to represent each interval.
(b) Construct a cumulative frequency table for these data

(c) Use the cumulative frequency table to draw, on graph paper, a cumulative
frequency graph, using a scale of 1 cm per 20 seconds waiting time for the
horizontal axis and 1 cm per 10 customers for the vertical axis.
(d) Use the cumulative frequency graph to find estimates for the median and the
lower and upper quartiles.

Ans.

(d) Median = 87 ± 2
Lower quartile = 65 ± 2
Upper quartile = 123 ± 2

### Question

[Maximum mark: 15] [with GDC]
In a suburb of a large city, 100 houses were sold in a three-month period. The following
cumulative frequency table shows the distribution of selling prices (in thousands of
dollars).

(a) Represent this information on a cumulative frequency curve, using a scale of 1
cm to represent $50000 on the horizontal axis, 1 cm to represent 5 houses on the vertical axis. (b) Use your curve to find the interquartile range. The information above is represented in the following frequency distribution. (c) Find the value of $$a$$ and of $$b$$. (d) Use mid-interval values to calculate an estimate for the mean selling price. (e) Houses which sell for more than$350000 are described as De Luxe.
(i) Use your graph to estimate the number of De Luxe houses sold.
(ii) Two De Luxe houses are selected at random. Find the probability that both
have a selling price of more than $400000. Answer/Explanation Ans. (a) (b) Q1 = 135 ± 5 Q3 = 240 ± 5 Interquartile range = 105 ± 10 (c) $$a$$ = 94 – 87 = 7, $$b$$ = 100 – 94 = 6 (d) mean = $$\frac{12(50)+46(150)+29(250)+7(350)+6(450)}{100}$$= 199 or$199000
OR directly by GDC

(e) (i) $350000 => 91.5 Number of De luxe houses $$\simeq$$ 100 – 91.5 = 9 or 8 (ii) P (both > 400000) = $$\frac{6}{9}\left ( \frac{5}{8} \right )=\frac{5}{12}or\frac{6}{8}\left ( \frac{5}{7} \right )=\frac{15}{28}$$ ### Question [Maximum mark: 16] [with GDC] One thousand candidates sit an examination. The distribution of marks is shown in the following grouped frequency table. (a) Complete the following table, which presents the above data as a cumulative frequency distribution. (b) Draw a cumulative frequency graph of the distribution, using a scale of 1 cm for 100 candidates on the vertical axis and 1 cm for 10 marks on the horizontal axis. (c) Use your graph to answer parts (i)–(iii) below, (i) Find an estimate for the median score. (ii) Candidates who scored less than 35 were required to retake the examination. How many candidates had to retake? (iii) The highest-scoring 15% of candidates were awarded a distinction. Find the mark above which a distinction was awarded. Answer/Explanation Ans. (c) (i) Median = 46 (ii) Scores < 35: 240 candidates (iii) Top 15% ⇒ Mark ≥ 63 ### Question [Maximum mark: 15] [with GDC] A supermarket records the amount of money d spent by customers in their store during a busy period. The results are as follows: (a) Find an estimate for the mean amount of money spent by the customers, giving your answer to the nearest dollar ($).
(b) Complete the following cumulative frequency table and use it to draw a
cumulative frequency graph. Use a scale of 2 cm to represent $20 on the horizontal axis, and 2 cm to represent 20 customers on the vertical axis. (c) The time $$t$$ (minutes), spent by customers in the store may be represented by the equation $$t=2d^{\frac{2}{3}}+3$$ (i) Use this equation and your answer to part (a) to estimate the mean time in minutes spent by customers in the store. (ii) Use the equation and the cumulative frequency graph to estimate the number of customers who spent more than 37 minutes in the store. Answer/Explanation Ans. (a) $$\bar{x}$$ =$59
OR

$$\bar{x}=\frac{10×24+30×16+…+110×10+130×4}{24+16+…+10+4}=\frac{7860}{134}= 59$$

(b)

(c) (i) $$t$$ = 2d2/3 + 3
Mean $$d$$= 59
Mean $$t$$ ≈ 2 × (59)2/3 + 3≈ 33.3 min. (3 sf) (accept 33.2)
(ii) $$t$$ > 37 ⇒ 2d2/3 + 3 > 37 ⇒ $$d$$ > 70.1
From the graph, when $$d$$ = 70.1, $$n$$ = 82
number of shoppers = 134 – 82 = 52

### Question

[Maximum mark: 10] [with GDC]
A taxi company has 200 taxi cabs. The cumulative frequency curve below shows the
fares in dollars ($) taken by the cabs on a particular morning. (a) Use the curve to estimate (i) the median fare; (ii) the number of cabs in which the fare taken is$35 or less.
The company charges 55 cents per kilometre for distance travelled. There are no other
charges. Use the curve to answer the following.
(b) On that morning, 40% of the cabs travel less than a km. Find the value of $$a$$.
(c) What percentage of the cabs travel more than 90 km on that morning?

Ans.

(a) (i) median fare = $$\$$24 (±0.5)
(ii) fare ≤ $$\$$35 => number of cabs is 154 (or 153)

(b) 40% of cabs = 80 cabs
fares up to $$\$$22
distance = $$\$$22 ÷ $0.55 $$a$$ = 40 km (c) Distance 90 km => fare = 90 × $$\$$0.55= $$\$$49.50 Fare$49.50 => number of cabs = 200 –186= 14
Thus percentage is $$\frac{14}{200}$$= 7%

### Question

[Maximum mark: 12] [with GDC]
There are 50 boxes in a factory. Their weights, $$w$$ kg, are divided into 5 classes, as
shown in the following table.

(a) Show that the estimated mean weight of the boxes is 32 kg.
(b) There are $$x$$ boxes in the factory marked “Fragile”. They are all in class E. The
estimated mean weight of all the other boxes in the factory is 30 kg. Calculate the
value of $$x$$.
(c) An additional $$y$$ boxes, all with a weight in class D, are delivered to the factory.
The total estimated mean weight of all of the boxes in the factory is less than 33
kg. Find the largest possible value of $$y$$.

Ans.

(a) mid interval values 14, 23, 32, 41, 50

$$\bar{w}=\frac{1600}{50}$$= 32 (kg)

(b) METHOD 1
Total weight of other boxes = 1600 – 50x
Total number of other boxes = 50 – x

$$\frac{1600-50x}{50-x}=30\Leftrightarrow 1600-50x=1500-30x\Leftrightarrow x=5$$

METHOD 2
Let $$z$$ be the number of other boxes in Class E
Total weight of other boxes = 1200 + 50z
Total number of other boxes = 42 + z

$$\frac{1200-50z}{42+z}=30\Leftrightarrow 1200+50z=1260+30z\Leftrightarrow z=3$$

x = 5

(c) $$\frac{98+276+416+41(10+y)+400}{50+y}< 33\Leftrightarrow \frac{1600-41y}{50+y}< 33$$
$$1600+41y<1650+33y\Leftrightarrow 8y<50$$
$$y < 6.25$$
6