Descriptive Statistics Practice Problems
1. The age in months at which a child first starts to walk is observed for a random group of children from a town in Brazil. The results are
14.3, 11.6, 12.2, 14.0, 20.4, 13.4, 12.9, 11.7, 13.1. (a) (i) Find the mean of the ages of these children.
(ii) Find the standard deviation of the ages of these children. (b) Find the median age.
(Total 6 marks)
2. The numbers of games played in each set of a tennis tournament were 9, 7, 8, 11, 9, 6, 10, 8, 12, 6, 8, 13, 7, 9, 10, 9, 10, 11, 12, 8, 7, 13, 10, 7, 7.
The raw data has been organized in the frequency table below. games frequency
6 2
7 5
8 n
9 4
10 4
11 2
12 2
13 2
(a) Write down the value of n.
(b) Calculate the mean number of games played per set. (c) What percentage of the sets had more than 10 games? (d) What is the modal number of games?
(Total 8 marks)
3. An atlas gives the following information about the approximate population of some cities in the year 2000. The population of Nairobi has accidentally been left out.
City Population in Millions
Melbourne 3.2
Bangkok 7.2
Nairobi
Paris 9.6
São Paulo 17.7
Tokyo 28.0
Seattle 2.1
The atlas tells us that the mean population for this group of cities is 10.01 million. (a) Calculate the population of Nairobi.
(b) Which city has the median population value?
(Total 8 marks)
4. In the following ordered data, the mean is 6 and the median is 5.2, b, 3, a, 6, 9, 10, 12 Find each of the following
(a) the value of a; (b) the value of b.
(Total 8 marks)
5. The weight in kilograms of 12 students in a class are as follows.63 76 99 65 63 51 52 95 63 71 65 83 (a) State the mode.
(b) Calculate
(i) the mean weight;
(ii) the standard deviation of the weights.
(2)
When one student leaves the class, the mean weight of the remaining 11 students becomes 70 kg. (c) Find the weight of the student who left.
(2)
(Total 5 marks)
6. Twenty students are asked how many detentions they received during the previous week at school. Theresults are summarised in the frequency distribution table below. Number of
detentions
x
Number of students
f
fx
0 6
1 3
2 10
3 1
Total 20
(a) What is the modal number of detentions received? (b) (i) Complete the table.
(ii) Find the mean number of detentions received.
(Total 4 marks)
7. For the set of {8, 4, 2, 10, 2, 5, 9, 12, 2, 6}(a) calculate the mean; (b) find the mode; (c) find the median.
(Total 4 marks)
8. The mean of the ten numbers listed below is 5.5.4, 3, a, 8, 7, 3, 9, 5, 8, 3 (a) Find the value of a.
(b) Find the median of these numbers.
(Total 4 marks)
9. The table shows the number of children in 50 families.Number of children
Frequency Cumulative frequency
1 3 3
2 m 22
3 12 34
4 p q
5 5 48
6 2 50
T
(a) Write down the value of T. (b) Find the values of m, p and q.
(Total 4 marks)
10. The temperatures in °C, at midday in Geneva, were measured for eight days and the results are recordedbelow.
7, 4, 5, 4, 8, T, 14, 4 The mean temperature was found to be 7 °C.
(3)
(b) Write down the mode.
(1)
(c) Find the median.
(2)
(Total 6 marks)
11. The grades obtained by a group of 20 IB students are listed below:6 2 5 3 5 5 6 2 6 1
7 6 2 4 2 4 3 4 5 6
(a) Complete the following table for the grades obtained by the students.
Grade Frequency
1 2
3 2
4
5 4
6
7 1
(2)
(b) Write down the modal grade obtained by the students.
(1)
(c) Calculate the median grade obtained by the students.
(2)
One student is chosen at random from the group.
(d) Find the probability that this student obtained either grade 4 or grade 5.
(1)
(Total 6 marks)
12. The figure below shows the lengths in centimetres of fish found in the net of a small trawler.11 10 9 8 7 6 5 4 3 2 1 –10 10 Number of
fish
20 30 40 50 60 70 80 90 110 120 130 Length (cm)
(a) Find the total number of fish in the net.
(2)
(b) Find (i) the modal length interval;
(ii) the interval containing the median length; (iii) an estimate of the mean length.
(5)
(c) (i) Write down an estimate for the standard deviation of the lengths.
(ii) How many fish (if any) have length greater than three standard deviations above the mean
?
The fishing company must pay a fine if more than 10% of the catch have lengths less than 40cm. (d) Do a calculation to decide whether the company is fined.
(2)
A sample of 15 of the fish was weighed. The weight, W was plotted against length, L as shown below. 1.2
1
0.8
0.6
0.4
0.2
W
(kg)
0 20 40 60 80 100
L (cm)
(e) Exactly two of the following statements about the plot could be correct. Identify the two correct statements.
(2)
Note: You do not need to enter data in a GDC or to calculate r exactly.(i) The value of r, the correlation coefficient, is approximately 0.871. (ii) There is an exact linear relation between W and L.
(iii) The line of regression of W on L has equation W = 0.012L + 0.008. (iv) There is negative correlation between the length and weight. (v) The value of r, the correlation coefficient, is approximately 0.998. (vi) The line of regression of W on L has equation W = 63.5L + 16.5.
(Total 14 marks)
13. A random sample of 167 people who own mobile phones was used to collect data on the amount of timethey spent per day using their phones. The results are displayed in the table below. Time spent per
day (t minutes) 0 ≤t<15 15 ≤t< 30 30 ≤t< 45 45 ≤t< 60 60 ≤t< 75 75 ≤t< 90
Number of people 21 32 35 41 27 11
(a) State the modal group.
(1)
(b) Use your graphic display calculator to calculate approximate values of the mean and standard deviation of the time spent per day on these mobile phones.
(3)
(c) On graph paper, draw a fully labelled histogram to represent the data.
(4)
(Total 8 marks)
14. The histogram below shows the amount of money spent on food each week by 45 families. Theamounts have been rounded to the nearest 10 dollars.
150 160 170 180 190
$
fr
eq
u
en
cy
18 16 14 12 10 8 6 4 2 0
(b) Find the largestpossibleamount spent on food by a single family in the modal group. (c) State which of the following amounts could not be the total spent by all families in the modal
group:
(i) $2430 (ii) $2495 (iii) $2500 (iv) $2520 (v) $2600
(Total 6 marks)
15. The heights in cm of the members of 4 volleyball teams A, B, C and D were taken and represented in thefrequency histograms given below. frequency
180
height (cm) A
190 200
frequency
180
height (cm) C
190 200
frequency
180
height (cm) B
190 200
frequency
180
height (cm) D
190 200
The mean x and standard deviation σ of each team are shown in the following table.
I II III IV
x 194 189 188 195
σ 6.50 4.91 3.90 3.74
Match each pair of x and σ (I, II, III, or IV) to the correct team (A, B, C or D).
x and σ Team I
II III IV
(Total 6 marks)
16. The following histogram shows the house prices in thousands of Australian dollars (AUD) of a randomsample of houses in a certain town in Australia.
(b) Write down the modal group for house prices.
(c) Find the probability of choosing a house at random that costs less than 60 000 AUD or more than 240 000 AUD.
(d) Given that a house costs more than 120 000 AUD, find the probability that it costs between 180 000 and 240 000 AUD.
(Total 6 marks)
17. The number of hours that a professional footballer trains each day in the month of June is represented in the following histogram.
10 9 8 7 6 5 4 3 2 1
0 1 2 3 4 5 6 7 8 9 10 number of hours
n
u
m
b
er
o
f
d
ay
s
(a) Write down the modal number of hours trained each day. (b) Calculate the mean number of hours he trains each day.
(Total 8 marks)
18. Fifty students at Layton High School recorded how much money each student in their class spent onvideo rentals this month (to the nearest dollar). The results are shown in the frequency table below:
Class interval in $ Boundaries in $ Frequency
1–10 0.50–10.50 10
11–20 10.50–20.50 20
21–30 20.50–30.50 10
31–40 30.50–40.50 0
41–50 40.50–50.50 4
51–60 50.50–60.50 2
61–70 60.50–70.50 4
(a) On graph paper using a scale of 2 cm to represent each interval ($10.00) on the horizontal axis and 1 cm to represent 5 people on the vertical axis, draw and clearly label a frequency histogram which displays the above information.
(5)
(b) Answer the following questions: (i) Which class is the modal class? (ii) In which class is the median?
(2)
(c) Assuming these students spend the same amount on videos each month, find the probability that next month a student will spend an amount in the class interval:
(i) From $21 to $30 inclusive on video rentals. (ii) $30 or less on video rentals.
(iii) From $41 to $60 on video rentals, given that they spent more than $20 on video rentals. (iv) Not more than $60 on video rentals, given that they spent over $10 on video rentals.
(6)
(Total 13 marks)
19. The bar chart below shows the number of people in a selection of families. 10
8 6 4 2 0
3 4 5 6 7 8 9 10
Number of people in a family Number of
families
(a) How many families are represented? (b) Write down the mode of the distribution.
(c) Find, correct to the nearest whole number, the mean number of people in a family.
(Total 4 marks)
20. The following table shows the age distribution of teachers who smoke at Laughlin High School.Ages Number of smokers 20 ≤x < 30 5
30 ≤x < 40 4 40 ≤x < 50 3 50 ≤x < 60 2 60 ≤x < 70 3 (a) Calculate an estimate of the mean smoking age.
(b) On the following grid, construct a histogram to represent this data.
(Total 4 marks)
21. A group of 25 females were asked how many children they each had. The results are shown in thehistogram below.
10 9 8 7 6 5 4 3 2 1 0
0 1 2 3 4
Number of Children Frequency
Number of Children per Female
(a) Show that the mean number of children per female is 1.4.
(2)
(b) Show clearly that the standard deviation for this data is approximately 1.06.
(3)
female was 2.4 and the standard deviation was 2. Use the results from parts (a) and (b) to describe the differences between the number of children the two groups of females have.
(2)
(d) A female is selected at random from the first group. What is the probability that she has more than two children?
(2)
(e) Two females are selected at random from the first group. What is the probability that (i) both females have more than two children?
(2)
(ii) only one of the females has more than two children?
(3)
(iii) the second female selected has two children given that the first female selected had no children?
(1)
(Total 15 marks)
22. David looked at a passage from a book. He recorded the number of words in each sentence as shown inthe following frequency table. Class interval (number of words)
Frequency
f
1–5 16
6–10 28
11–15 26
16–20 14
21–25 10
26–30 3
31–35 1
36–40 0
41–45 2
(a) Find the class interval in which the median lies.
(b) Estimate, correct to the nearest whole number, the mean number of words in a sentence.
(Total 4 marks)
23. There are 120 teachers in a school. Their ages are represented by the cumulative frequency graph below.0 5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 Age
130 120 110 100 90 80 70 60 50 40 30 20 10 0
C
u
m
u
la
ti
v
e
fr
eq
u
en
(a) Write down the median age.
(1)
(b) Find the interquartile range for the ages.
(2)
(c) Given that the youngest teacher is 21 years old and the oldest is 72 years old, represent the information on a box and whisker plot using the scale below.
0 5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 Age
(3)
(Total 6 marks)
24. A random sample of 200 females measured the length of their hair in cm. The results are displayed in thecumulative frequency curve below.
200 175 150 125 100 75 50 25 0
50 45 40 35 30 25 20 15 10 5 0
Cu
m
u
la
ti
v
e
fr
eq
u
en
cy
length (cm)
(a) Write down the median length of hair in the sample.
(1)
(b) Find the interquartile range for the length of hair in the sample.
(2)
(c) Given that the shortest length was 6 cm and the longest 47 cm, draw and label a box and whisker plot for the data on the grid provided below.
50 45 40 35 30 25 20 15 10 5 0
length (cm)
(3)
(Total 6 marks)
25. The diagram below shows the cumulative frequency distribution of the heights in metres of 600 trees in a(a) Write down the median height of the trees.
(1)
(b) Calculate the interquartile range of the heights of the trees.
(2)
(c) Given that the smallest tree in the wood is 3 m high and the tallest tree is 28 m high, draw the box and whisker plot on the grid below that shows the distribution of trees in the wood.
(3)
(Total 6 marks)
26. (a) State which of the following sets of data are discrete.(i) Speeds of cars travelling along a road. (ii) Numbers of members in families. (iii) Maximum daily temperatures.
(iv) Heights of people in a class measured to the nearest cm. (v) Daily intake of protein by members of a sporting team. The boxplot below shows the statistics for a set of data.
2 4 6 8 10 12 14 16 18 20 data values
(b) For this data set write down the value of (i) the median;
(ii) the upper quartile;
(iii) the minimum value present.
(c) Write down three different integers whose mean is 10.
(Total 6 marks)
27. (a) The exam results for 100 boys are displayed in the following diagram:0 10 20 30 40 50 60 70 80 90 100
(i) Find the range of the results. (ii) Find the interquartile range. (iii) Write down the median.
(b) The exam results for 100 girls are displayed in the diagram below: 100
90
80
70
60
50
40
30
20
10
0
10 20 30 40 50 60 70 80 90 100 exam results
n
u
m
b
er
o
f
g
ir
ls
c
u
m
u
la
ti
v
e
fr
eq
u
en
cy
(i) Write down the median. (ii) Find the inter quartile range.
(c) Write down the set of results that are the most spread out and give a reason for your answer.
(Total 6 marks)
28. The local council has been monitoring the number of cars parked near a supermarket on an hourly basis. The results are displayed below.
Parked Cars/Hour Frequency Cumulative Frequency
0–19 3 3
20–39 15 18
40–59 25 w
60–79 35 78
80–99 17 95
(a) Write down the value of w.
(b) Draw and label the Cumulative Frequency graph for this data.
(c) Determine the median number of cars per hour parked near the supermarket.
(Total 8 marks)
29. The cumulative frequency table below shows the ages of 200 students at a college. Age Number of Students Cumulative Frequency
17 3 3
18 72 75
19 62 137
20 31 m
21 12 180
22 9 189
23–25 5 194
> 25 6 n
(a) What are the values of m and n?
(b) How many students are younger than 20? (c) Find the value in years of the lower quartile.
(Total 8 marks)
30. The table below shows the number and weight (w) of fish delivered to a local fish market one morning.weight (kg) frequency cumulative frequency
0.70 ≤w < 0.90 37 53
0.90 ≤w < 1.10 44 c
1.10 ≤w < 1.30 23 120
1.30 ≤w < 1.50 10 130
(a) (i) Write down the value of c.
(1)
(ii) On graph paper, draw the cumulative frequency curve for this data. Use a scale of 1 cm to represent 0.1 kg on the horizontal axis and 1 cm to represent 10 units on the vertical axis. Label the axes clearly.
(4)
(iii) Use the graph to show that the median weight of the fish is 0.95 kg.
(1)
(b) (i) The zoo buys all fish whose weights are above the 90th percentile. How many fish does the zoo buy?
(2)
(ii) A pet food company buys all the fish in the lowest quartile. What is the maximum weight of a fish bought by the company?
(3)
(c) A restaurant buys all fish whose weights are within 10% of the median weight.
(i) Calculate the minimum and maximum weights for the fish bought by the restaurant.
(2)
(ii) Use your graph to determine how many fish will be bought by the restaurant.
(3)
(Total 16 marks)
31. The cumulative frequency graph has been drawn from a frequency table showing the time it takes anumber of students to complete a computer game. 200
180 160 140 120 100 80 60 40 20
0 5 10 15 20 25 30 35 40 45 50 55 60
Time in minutes
N
u
m
b
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o
f
st
u
d
en
ts
f
(a) From the graph find (i) the median time; (ii) the interquartile range.
The graph has been drawn from the data given in the table below.
Time in minutes Number of students
0 < x≤ 5 20 5 < x≤ 15 20
15 < x≤ 20 p
20 < x≤ 25 40
25 < x≤ 35 60
35 < x≤ 50 q
50 < x≤ 60 10
(b) Using the graph, find the values of p and q.
(2)
(c) Calculate an estimate of the mean time taken to finish the computer game.
(4)
(Total 11 marks)
32. The heights of 200 students are recorded in the following table.Height (h) in cm Frequency 140 ≤h < 150 2
150 ≤h < 160 28
160 ≤h < 170 63 170 ≤h < 180 74 180 ≤h < 190 20 190 ≤h < 200 11 200 ≤h < 210 2 (a) Write down the modal group.
(1)
(b) Calculate an estimate of the mean and standard deviation of the heights.
(4)
The cumulative frequency curve for this data is drawn below. 200
180 160 140 120 100 80 60 40 20 0
140 150 160 170 180 190 200 210
height in cm
n
u
m
b
er
o
f
st
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d
en
ts
(c) Write down the median height.
(d) The upper quartile is 177.3 cm. Calculate the interquartile range.
(2)
(e) Find the percentage of students with heights less than 165 cm.
(2)
(Total 10 marks)
33. The cumulative frequency graph below shows the examination scores of 80 students.80 70 60 50 40 30 20 10
10 20 30 40 50 60 scores
cumulative frequency
0 From the graph find
(a) the median value; (b) the interquartile range; (c) the 35th percentile;
(d) the percentage of students who scored 50 or above on this examination.
(Total 8 marks)
34. The graph below shows the cumulative frequency for the yearly incomes of 200 people.200
180
160
140
120
100
80
60
40
20
0
0 5000 10 000 15 000 20 000 25 000 30 000 35 000 Annual income in British pounds
Cumulative frequency
Use the graph to estimate
(a) the number of people who earn less than 5000 British pounds per year; (b) the median salary of the group of 200 people;
(c) the lowest income of the richest 20% of this group.
35. The table below shows the percentage, to the nearest whole number, scored by candidates in an examination.
Marks (%) 0–9 10–19 20–29 30–39 40–49 50–59 60–69 70–79 80–89 90–100
Frequency 2 7 8 13 24 30 6 5 3 2
The following is the cumulative frequency table for the marks. Marks (%) Cumulative frequency
< 9.5 2 < 19.5 9 < 29.5 s
< 39.5 30 < 49.5 54 < 59.5 84 < 69.5 t
< 79.5 95 < 89.5 98
< 100 100
(a) Calculate the values of s and of t.
(2)
(b) Using a scale of 1 cm to represent 10 marks on the horizontal axis, and 1 cm to represent 10 candidates on the vertical axis, draw a cumulative frequency graph.
(3)
(c) Use your graph to estimate (i) the median mark; (ii) the lower quartile;
(iii) the pass mark, if 40% of the candidates passed.
(4)
(Total 9 marks)
36. The following table shows the times, to the nearest minute, taken by 100 students to complete amathematics task.
Time (t) minutes 11–15 16–20 21–25 26–30 31–35 36–40
Number of students 7 13 25 28 20 7
(a) Construct a cumulative frequency table. (Use upper class boundaries 15.5, 20.5 and so on.)
(2)
(b) On graph paper, draw a cumulative frequency graph, using a scale of 2 cm to represent 5 minutes on the horizontal axis and 1 cm to represent 10 students on the vertical axis.
(3)
(c) Use your graph to estimate
(i) the number of students that completed the task in less than 17.5 minutes; (ii) the time it will take for
4 3
of the students to complete the task.
(2)
(Total 7 marks)
37. A marine biologist records as a frequency distribution the lengths (L), measured to the nearestcentimetre, of 100 mackerel. The results are given in the table below. Length of mackerel
(L cm)
Number of mackerel 27 < L≤ 29 2
29 < L ≤ 31 4 31 < L ≤ 33 8 33 < L ≤ 35 21
35 < L ≤ 37 30 37 < L ≤ 39 18
39 < L ≤ 41 12
41 < L ≤ 43 5 100
(a) Construct a cumulative frequency table for the data in the table.
(2)
(b) Draw a cumulative frequency curve.
Hint: Plot your cumulative frequencies at the top of each interval.
(3)
(c) Use the cumulative frequency curve to find an estimate, to the nearest cm for (i) the median length of mackerel;
(2)
(ii) the interquartile range of mackerel length.
(2)
(Total 9 marks)
38. The following histogram shows the weights of a number of frozen chickens in a supermarket. Theweights are grouped such that 1 ≤ weight < 2, 2 ≤ weight < 3 and so on.
55 50 45 40 35 30 25 20 15 10 5 0
6 5
4 3
2 1
0 number of
chickens
weight (kg)
(a) On the graph above, draw in the frequency polygon.
(2)
(b) Find the total number of chickens.
(1)
(c) Write down the modal group.
(1)
Gabriel chooses a chicken at random.
(d) Find the probability that this chicken weighs less than 4 kg.
(2)
(Total 6 marks)
39. A survey was conducted of the number of bedrooms in 208 randomly chosen houses. The results areshown in the following table.
Number of bedrooms 1 2 3 4 5 6
Number of houses 41 60 52 32 15 8
(a) State whether the data is discrete or continuous.
(1)
(2)
(c) Write down the standard deviation of the number of bedrooms per house.
(1)
(d) Find how many houses have a number of bedrooms greater than one standard deviation above the mean.
(2)
(Total 6 marks)
40. The number of bottles of water sold at a railway station on each day is given in the following table.Day 0 1 2 3 4 5 6 7 8 9 10 11 12
Temperature
(T°) 21 20.7 20 19 18 17.3 17 17.3 18 19 20 20.7 21 Number of
bottles sold (n) 150 141 126 125 98 101 93 99 116 121 119 134 141 (a) Write down
(i) the mean temperature;
(ii) the standard deviation of the temperatures.
(2)
(b) Write down the correlation coefficient, r, for the variables n and T.
(1)
(c) Comment on your value for r.
(2)
(d) The equation of the line of regression for n on T is n = dT −100. (i) Write down the value of d.
(ii) Estimate how many bottles of water will be sold when the temperature is 19.6°.
(2)
(e) On a day when the temperature was 36° Peter calculates that 314 bottles would be sold. Give one reason why his answer might be unreliable.
(1)
(Total 8 marks)
41. The distribution of the weights, correct to the nearest kilogram, of the members of a football club isshown in the following table.
Weight (kg) 40 – 49 50 – 59 60 – 69 70 – 79
Frequency 6 18 14 4
(a) On the grid below draw a histogram to show the above weight distribution.
(2)
(b) Write down the mid-interval value for the 40 – 49 interval.
(1)
(2)
(d) Write down an estimate of the standard deviation of their weights.
(1)
(Total 6 marks)
42. Five pipes labelled, “6 metres in length”, were delivered to a building site. The contractor measured eachpipe to check its length (in metres) and recorded the following; 5.96, 5.95, 6.02, 5.95, 5.99. (a) (i) Find the mean of the contractor’s measurements.
(ii) Calculate the percentage error between the mean and the stated, approximate length of 6 metres.
(b) Calculate 3.895−8.73−0.5 , giving your answer (i) correct to the nearest integer;
(ii) in the form a ×10k, where 1 ≤a< 10, k ∈ .
(Total 6 marks)
43. (a) Write down the following numbers in increasing order.3.5, 1.6 ×10−19, 60730, 6.073×105, 0.006073×106, π, 9.8×10−18. (b) Write down the median of the numbers in part (a).
(c) State which of the numbers in part (a) is irrational.
(Total 6 marks)
44. In an experiment a vertical spring was fixed at its upper end. It was stretched by hanging differentweights on its lower end. The length of the spring was then measured. The following readings were obtained.
Load (kg) x 0 1 2 3 4 5 6 7 8
Length (cm) y 23.5 25 26.5 27 28.5 31.5 34.5 36 37.5 (a) Plot these pairs of values on a scatter diagram taking 1 cm to represent 1 kg on the horizontal axis
and 1 cm to represent 2 cm on the vertical axis.
(4)
(b) (i) Write down the mean value of the load (x).
(1)
(ii) Write down the standard deviation of the load.
(1)
(iii) Write down the mean value of the length (
y
).(1)
(iv) Write down the standard deviation of the length.
(1)
(c) Plot the mean point (x,
y
) on the scatter diagram. Name it L.(1)
It is given that the covariance Sxy is 12.17.
(d) (i) Write down the correlation coefficient, r, for these readings.
(1)
(ii) Comment on this result.
(2)
(e) Find the equation of the regression line of y on x.
(2)
(f) Draw the line of regression on the scatter diagram.
(2)
(g) (i) Using your diagram or otherwise, estimate the length of the spring when a load of 5.4 kg is applied.
(1)
(ii) Malcolm uses the equation to claim that a weight of 30 kg would result in a length of 62.8 cm. Comment on his claim.
(1)
(Total 18 marks)
45. Peter has marked 80 exam scripts. He has calculated the mean mark for the scripts to be 62.1. Maria hasmarked 60 scripts with a mean mark of 56.8.
(a) Peter discovers an error in his marking. He gives two extra marks each to eleven of the scripts. Calculate the new value of the mean for Peter’s scripts.
(b) After the corrections have been made and the marks changed, Peter and Maria put all their scripts together. Calculate the value of the mean for all the scripts.
(Total 8 marks)
46. A group of students has measured the heights of 90 trees. The class calculate the mean height to be x = 12.4 m with standard deviation s = 5.35 m. One student notices that two of the measurements, 44.5 m and 43.2 m, are much too big and must be wrong.
(a) How many standard deviations away from the mean of 12.4 is the value 44.5? The incorrect measurements of 44.5 m and 43.2 m must be removed from the data. (b) Calculate the new value of x after removing the two unwanted values.
(Total 8 marks)
47. A shopkeeper wanted to investigate whether or not there was a correlation between the prices of food 10years ago in 1992, with their prices today. He chose 8 everyday items and the prices are given in the table below.
sugar milk eggs rolls tea bags coffee potatoes flour 1992 price $1.44 $0.80 $2.16 $1.80 $0.92 $3.16 $1.32 $1.12 2002 price $2.20 $1.04 $2.64 $3.00 $1.32 $2.28 $1.92 $1.44
(a) Calculate the mean and the standard deviation of the prices (i) in 1992;
(ii) in 2002.
(4)
(b) (i) Given that sxy = 0.3104, calculate the correlation coefficient.
(ii) Comment on the relationship between the prices.
(4)
(c) Find the equation of the line of the best fit in the form y = mx + c.
(3)
(d) What would you expect to pay now for an item costing $2.60 in 1992?
(1)
(e) Which item would you omit to increase the correlation coefficient?
(2)
(Total 14 marks)
48. The heights and weights of 10 students selected at random are shown in the table below.Student 1 2 3 4 5 6 7 8 9 10
Height
x cm
155 161 173 150 182 165 170 185 175 145 Weight
y kg
50 75 80 46 81 79 64 92 74 108
(a) Plot this information on a scatter graph. Use a scale of 1 cm to represent 20 cm on the
x-axis and 1 cm to represent 10 kg on the y-axis.
(4)
(b) Calculate the mean height.
(1)
(c) Calculate the mean weight.
(1)
(d) It is given that Sxy = 44.31.
show that the gradient of the line of regression of y on x is 0.276. (ii) Calculate the equation of the line of best fit.
(iii) Draw the line of best fit on your graph.
(6)
(e) Use your line to estimate
(i) the weight of a student of height 190 cm; (ii) the height of a student of weight 72 kg.
(2)
(f) It is decided to remove the data for student number 10 from all calculations. Explain briefly what effect this will have on the line of best fit.
(1)
(Total 15 marks)
49. The following table gives the heights and weights of five sixteen-year-old boys.Name Height Weight Blake 182 cm 73 kg
Jorge 173 cm 68 kg Chin 162 cm 60 kg Ravi 178 cm 66 kg Derek 190 cm 75 kg (a) Find
(i) the mean height; (ii) the mean weight.
(b) Plot the above data on the grid below and draw the line of best fit. 190
185 180 175 170 165 160
0
60 65 70 75
weight (kg) height
(cm)
Descriptive Statistics Practice Problems: MarkScheme
1. (a) (i) mean = 13.7 (M1)(A1) (G2)
(ii) sd = 2.52 (M1)(A1) (G2)
(b) For attempting to put their numbers in order (M1)
13.1 (A1) (G2)
[6]
2. (a) n = 4 (A2) (C2)
(b) Mean number of games is 9.08 (accept 9). (M1)(A1) (C2)
Note:Award(M1)forindicatingasumofgamestimesfrequency(possibly curtailedbydots)orfor227seen.
(c)
1 100 25
6 ×
= 24% (M1)(A1) (C2)
Note:Award(M1)(A0)if6isreplacedby10.Nootheralternative.
(d) Modal number of games is 7. (A2) (C2)
[8] 3. (a) µ =
7 1 . 2 0 . 28 7 . 17 6 . 9 2 . 7 2 .
3 + +x+ + + +
= 10.01. (M1)
Hence 67.8 + x = 10.01 × 7 = 70.07. (A1)(M1) x = 70.07 – 67.8 = 2.27 (accept 2.27 or 2.3 million). (M1)(A1) (C5) (b) Median is the middle value which is 7.2. (M1)(A1)
Bangkok. (A1) (C3)
Note: Award (M0)(A0)(A0) for Paris.
[8] 4. (a) 2 6 + a
= 5 (M1)(A1)
a + 6 = 10 (A1)
a = 4 (A1) (C4)
(b)
8 42+a+b
= 6 (M1)
42 + a + b = 48 (A1)
a + b = 6
4 + b = 6 (A1)
b = 2 (A1) (C4)
[8]
5. (a) 63 kg (A1) 1
(b) (i) 70.5 kg (G1)
(ii) 14.6 kg (also accept 15.2 kg) (G1) 2 (c) Total weight of 12 students = 846 kg
Total weight of 11 students = 11 × 70 = 770 kg (M1) Weight of student who left = 846 – 770 = 76 kg (A1) 2
[5]
6. (a) Mode = 2 (A1)
(b) (i)
x f fx
0 1 2 3 6 3 10 1 0 3 20 3
Total 20 26 (A1)
Note: Award (A1) for three or more correct bold entries. (ii) Mean =
20 26
(M1)
Note: Award (M1) for dividing fx total by 20.
= 1.3 (A1)
Mean = 1.3 (C2)
[4] 7. (a) Mean =
10 60
= 6 (A1) (C1)
(b) Mode = 2 (A1) (C1)
(c) 2, 2, 2, 4, 5, 6, 8, 9, 10, 12 Median =
2 6 5+
↑
(M1)
= 5.5 (A1) (C2)
[4] 8. (a) 5.5 =
10
3 8 5 9 3 7 8 3
4+ +a+ + + + + + +
(M1) 55 = 50 + a
5 = a (A1) (C2)
(b) 3, 3, 3, 4, 5, 5, 7, 8, 8, 9 (M1)
Median = 5 (A1) (C2)
Note: Award (M1) for arranging scores in ascending or descending order. Follow through with candidate’s a
[4]
9. (a) T = 50 (A1)
(b) m = 19 (A1)
(c) p = 9 (A1)
(d) q = 43 (A1)
[4]
10. (a) 7
8
4 14 8
4 5 4
7+ + + + +T+ + =
(A1)(A1)
Note: Award (A1) for sum +T, (A1) for 56 or 7 × 8 or 8 in the denominator and 7 seen.
T= 10 (A1) (C3)
(b) 4 (A1) (C1)
(c) 4, 4, 4, 5, 7, 8, 10, 14 (M1)
Note: Award (M1) for arranging their numbers in order.
Median = 6 (A1)(ft) (C2)
[6] 11. (a)
Grade Frequency
1 1
2 4
3 (2)
4 3
5 (4)
6 5
7 (1)
(A2) (C2)
Note: Award (A1) for three correct. Award (A0) for two or fewer correct.
(b) Mode = 6 (A1)(ft) (C1)
(c) Median = 4.5 (M1)
Note: (M1) for attempt to order raw data (if frequency table
not used) (A1)(ft)
(d)
20 7
(0.35, 35 %) (A1)(ft) (C1)
[6] 12. Unit penalty (UP) is applicable where indicated.
(a) Total = 2 + 3 + 5 + 7 + 11 + 5 + 6 + 9 + 2 + 1 (M1)
Note: (M1) is for a sum of frequencies.
= 51 (A1)(G2) 2
(b) (i) modal interval is 60 – 70 (A1)
Note: Award (A0) for 65
(ii) median is length of fish no. 26, (M1)
also 60 – 70 (A1)(G2)
Note: Can award (A1)(ft) or (G2)(ft) for 65 if (A0) was awarded for 65 in part (i).
UP (iii) mean is
51
... 55 7 45 5 35 3 25
2× + × + × + × +
(M1)
UP = 69.5 cm (3s.f.) (A1)(ft)(G1) 5
Note: (M1) is for a sum of (frequencies multiplied by mid-point values) divided by candidate’s answer from part (a). Accept mid-points 25.5, 35.5 etc or 24.5, 34.5 etc, leading to answers 70.0 or 69.0 (3s.f.) respectively. Answers of 69.0, 69.5 or 70.0 (3s.f.) with no working can be awarded (G1).
UP (c) (i) standard deviation is 21.8 cm (G1)
Note: For any other answer without working, award (G0). If working is present then (G0)(AP) is possible.
(ii) 69.5 + 3 × 21.8 = 134.9 > 120 (M1)
no fish (A1)(ft)(G1) 3
Note: For ‘no fish’ without working, award (G1) regardless of answer to (c)(i). Follow through from (c)(i) only if method is shown.
(d) 5 fish are less than 40 cm in length, (M1) Award (M1) for any of ,
51 46 , 51
5
0.098 or 9.8%, 0.902, 90.2% or 5.1 seen.
hence no fine. (A1)(ft) 2
Note: There is no G mark here and (M0)(A1) is never allowed. The follow-through is from answer in part (a).
(e) (i) and (iii) are correct. (A1)(A1) 2
[14] 13. Unit penalty (UP) is applicable in question part (b) only.
(a) 45 ≤t < 60 (A1) 1
(b) 42.4 minutes (G2)
UP 21.6 minutes (G1) 3
(c)
(A4) 4
[8] 14. (a) mean =
45
190) 3 180 7 170 11 160 16 150
(8× + × + × + × + ×
(M1)(M1)
Note: Award (M1) for five correct products shown or implied in the numerator, (M1) for denominator 45.
= $165.78 per week (allow $166) (A1) (C3)
Notes: For165.7ɺ or
9 7
165 award (C3) for exact answer. For 165.77 award (C2) and no (AP).
For 165.77 with no working award (C2)(A0)(AP).
(b) $164.99 ($165) (A1) (C1)
(c) 16 × $155 = $2480 and 16 × 164.99 = $2639.84 ($2640) (M1)
Note: The (M1) is for a sensible attempt to calculate both bounds or for showing division by 16 of any of the values (i) to (v).
$2430 is not possible (A1)(ft) (C2)
Note: Follow through if wrong modal group is used in (b).
[6] 15.
x
and σ TeamI B
II C
III D
IV A
(A6) (C6)
Note: Award (A6) for all correct, (A4) for 2 correct or for 3 correct and 1 blank, (A2) for 1 correct but (A0) if the same letter appears 4 times.
[6]
16. (a) 109 (A1) (C1)
(b) 60–120 thousand dollars (A1) (C1)
(c)
109 32
For correct numerator (A1)
For correct denominator (A1) (C2)
(d)
39 10
For correct numerator (A1)
For correct denominator (A1) (C2)
[6]
17. (a) 6 hours (accept (5.5–6.5)) (A2) (C2)
(b)
30
) 4 8 8 7 9 6 5 5 4 4
( × + × + × + × + ×
(M1)(A2)(A1) =
30 183
= 6.1 (A2) (C6)
Note: Award (M1) for method, (A2) for all 5 terms in numerator correct. ((A1) for 3 or 4 terms in the numerator correct), (A1) for denominator.
[8] 18. (a)
25 20 15 10 5 0
0.50 10.50 20.50 30.50 40.50 50.50 60.50 70.50 Dollars (class boundaries)
N
u
m
b
er
o
f
p
eo
p
le
(
fr
eq
u
en
cy
)
(A5) 5
Notes: Award (A1) for correct scales, (A1) for bars connected (where possible), (A1) for axes labelled correctly, (A2) for bars all correctly drawn.
If one bar is incorrect, award (A1), if 2 or more bars are incorrect, award (A0). 30.50–40.50 counts as a bar, height zero. A1so, allow horizontal scale to be 10, 20, 30, etc, with bars moved 0.50 to the right.
(b) (i) 11–20 (accept 10.50–20.50) (A1)
(ii) 11–20 (accept 10.50–20.50) (A1) 2
(c) (i)
5 1 50 10=
or 20% (A1)
(ii)
5 4 50 40=
or 80% (A1)
(iii)
10 3 20
6 =
or 30% (A2)
Note: Award (A1) for correct numerator and (A1) for correct denominator. (iv)
10 9 40 36=
or 90% (A2) 6
Note: Award (A1) for correct numerator and (A1) for correct denominator.
[13]
19. (a) 30 (A1)
(b) 6 (A1)
(c)
30 1
((3 × 2) + (4 × 4) + ... + (10 × 1)) = 5.9 (M1)
= 6 (nearest whole number) (A1)
[4] 20. (a) mean =
17 3 ) 65 ( 2 ) 55 ( 3 ) 45 ( 4 ) 35 ( 5 ) 25 ( + + + + (M1)
Note: Award (M1) for using mid-interval values.
mean = 41.5 (A1)
(b) 5 4 3 2 1
20 30 40 50 60 70
ages frequency
(# of smokers)
(A2)
Note: Award (A1) for correct intervals, (A1) for correct bar lengths
[4] 21. (a) Mean =
25 4 1 3 3 2 6 1 10 0
5× + × + × + × + ×
= 1.4 (M2)(AG) 2
Note: Award (M1) for the numerator and (M1) for the denominator. (b)
∑
− 2_
) (x x
f = 5(0 – 1.4)2 + 10(1 – 1.4)2 + 6(2 – 1.4)2
+ 3(3 – 1.4)2 + 1(4 – 1.4)2 = 28 (M2)
Note: Award (M1) for 2
_
)
(x−x values, and (M1) for multiplying by the appropriate frequencies.
S.D. =
25 28
(M1)
= 1.06 (AG) 3
(c) Award (R1) for each acceptable reason, e.g. Group 2 has more children in total.
Group 2 has a larger number of children per female.
Group 2 generally have larger families. (R2) 2 (d) P(> 2 children) =
25 1 3+
(M1) =
25 4
(A1) 2 (e) (i) P(both females have > 2 children) =
24 3 25
4 ×
(M1) =
600 12
or
50 1
or 0.02 (A1)
(ii) P(only 1 female has > 2 children) = 2 ×
24 21 25
4 ×
(M2)
Note: Award (M1) for
24 21 25
4 ×
, (M1) for multiplying by 2.
=
600 168
or
75 21
or 0.28 (A1)
(iii) P(second has 2 childrenfirst has 0) =
24 6
or
4 1
or 0.25 (A1) 6
[15]
22. (a) Interval 11–15 (A1)
(b) Mid-intervals 3, 8, 13, 18 ... (M1)
Note: Award (M1) for all correct numbers.
Σxf = 48 + 224 + 338 + ... (M1)
Note: Award (M1) for attempt to obtain sum.
Mean = 13 (A1)
[4]
23. (a) Median = 45 (A1)
Note: Accept 45.5 (C1)
(b) 53 – 37 for identifying correct quartiles (A1) = 16 for correct answer to subtraction (A1)(ft)
Note: (ft)on their quartiles (C2)
(c)
Median marked correctly. (A1)(ft)
Box with ends at candidate’s quartiles. (A1)(ft) End points at 21 and 72 joined to box with straight lines. (A1)
Note: Award (A0) if lines go right through the box. (C3)
[6] 24. Unit penalty (UP) is applicable where indicated.
UP (a) 26cm (A1) (C1)
UP =14cm. (A1)(ft)
(ft) on their quartiles. (C2)
(c)
50 45 40 35 30 25 20 15 10 5 0
length (cm)
correct median (A1)(ft)
correct quartiles and box (A1)(ft)
endpoints at 6 and 47, joined to box by straight lines. (A1) (C3)
[6] 25. Unitpenalty(UP)appliesinpart(a)inthisquestion
(a) Median = 11m (A1) (C1)
(b) Interquartile range = 14 – 10 (A1)
= 4 (A1)(ft) (C2)
Note: (M1) for taking a sensible difference or for both correct quartile values seen.
(c)
correct median (A1)(ft)
correct quartiles and box (A1)(ft)
endpoints at 3 and 28, joined to box by straight lines (A1) (C3)
Note: Award (A0) if the lines go right through the box. Award final (A1) if the whisker goes to 20 with an outlier at 28
[6] 26. (a) (ii) and (iv) are discrete. (A1)(A1) (C2)
Notes: Award (A1)(A0) for both correct and one incorrect. Award (A1)(A0) for one correct and two incorrect. Otherwise, (A0)(A0).
(b) (i) Median = 10 (A1)
(ii) Q3 = 12 (A1)
(iii) Min value = 1(±0.2) (A1) (C3)
(c) Any three different integers whose mean is 10 eg 9, 10, 11. (A1) (C1)
[6]
27. (a) (i) 95 – 6 = 89 (A1)
(ii) 73 – 50 = 23 (A1)
(iii) 60 (A1) (C3)
(b) (i) 62 (A1)
(ii) 73 – 43 = 30 (A1) (C2)
(c) The girls as the IQR is larger (R1) (C1)
[6]
28. (a) w = 43 (C1)
100
80
60
40
20
0 20 40 60 80 100
number of cars
cu
m
u
la
ti
v
e
fr
eq
u
en
cy
cars parked per hour
(A5) (C5)
Notes: Award (A1) for labels (“cumulative” is not essential). Award (A1) for scales.
Award (A2) for all points correct, (A1)(A0) for 4 points correct. Award (A1) for neat curve or straight line segments.
The curve must extend to zero for the last (A1).
The points must be at 19, 39, etc (ft from candidate’s scale).
If the points are displaced consistently, (eg to 20,40 etc or to mid-points) then award (A0)(A1) ft for points.
If the scale is marked as 19,39, etc, this is allowed as long as 0 is not included, is shown displaced right by one unit.
Scale marked as an interval (eg 0−19) is acceptable, if written within the interval.
Bar graph or histogram receive (A0) for curve but if bars are at correct height and terminate at 19, 39, etc or have correct points marked on them, then award (A2) for points or (A0)(A1) ft if points are consistently incorrect.
(c) median = 63 ± 2 (M1)(A1) (C2)
Note: Award (M1) for a horizontal line drawn at height 48. Answer must be an integer, follow through from candidate’s graph.
[8]
29. (a) m = 137 + 31 = 168, (M1)(A1)
n = 194 + 6 = 200 or just n = 200 (M1)(A1) (C4)
(b) 137 students are aged below 20. (A2) (C2)
(c) 25% of 200 is 50. (M1)
50th student is 18 years old. (A1) (C2)
[8]
30. (a) (i) c = 97 (A1) 1
(ii) (not drawn to scale)
Weights of fish
cu
m
u
la
ti
v
e
fr
eq
u
en
cy
140 120 100 80 60 40 20 0
0.5 0.7 0.9 1.1 1.3 1.5
weight (kg) (A4) 4
(A1) for 3 to 4 correct, (A1) for the curve.
(iii) median is 0.95 kg line drawn correctly on diagram (M1)(AG) 1 (b) (i) 90th percentile ... 130 × 0.9 = 117 (M1)
The zoo buys 13 fish (±2). (A1) or (G2) 2
(ii) First quartile = 32/33 fish. (M1)
Maximum weight = 0.79 kg (±0.03). (M1)(A1) 3 (c) (i) maximum : 0.95 × 1.10 = 1.045 kg (1.05 to 3 s.f.) (A1)
minimum : 0.95 × 0.90 = 0.855 kg (A1) 2
(ii) number of fish bought by restaurant = 88 – 46 = 42(±4) (M1)(M1)(A1) 3
[16]
31. (a) (i) 25 minutes (±2 minutes) (A2)
(ii) Lower quartile = 18 (±1 minute) (A1) Upper quartile = 32 (±1 minute) (A1) Interquartile range = 32 – 18 = 14 minutes (±2 minutes) (A1)
OR
Accept [18 to 32] as interval for the interquartile range. (A3) 5
(b) p = 20 (A1)
q = 30 (A1) 2
(c)
Midpoint Frequency M × f
2.5 20 50
10 20 200
17.5 20 350
22.5 40 900
30 60 1800
42.5 30 1275
55 10 550
(A1) Total = 200 Total = 5125 (A1) Mean =
200 5125
= 25.625 (exact) or 25.6 (3 s.f.) (M1)(A1)
Note: Not every step needs to be seen to get the marks. OR
Mean = 25.625 or 25.6 (using GDC) (G4) 4
[11] 32. (a) Modal group =170 ≤h < 180 (A1) 1
(b) Mean = 171 (G2)
Standard deviation = 11.1 (G2) 4
(c) Median = 171 (±1) (A1) 1
(d) Lower quartile = 164.5 (±1) (A1)
Inter-quartile range = 177.3 – 164.5 = 12.8 (A1) 2
(e) number = 52 (±2) (A1)
percentage =
200 52
× 100 = 26% (A1) 2
[10] 33.
80 70 60 50 40 30 20 10
10 20 30 40 50 60 70 scores
cumulative frequency
0
(a) 30 (A2) (C2)
(b) P75 = 46, P25 = 20 (M1)
46 – 20 = 26 (A1) (C2)
(c) 0.35(80) = 28, P35 = 23 (A2) (C2)
(d)
80 10
= 12.5% (A2) (C2)
Note: Allow ±2 for each measurement.
Note: Allow ft for (b), (c) and (d) if percentile scores were figured on the basis of 100 instead of 80.
[8]
34. (a) 19 or 20 people (A1)
(b) Median salary = 15000 GBP (A1)
(c) 80% of 200 = 160
23000 ± 500 (M1)
(A1)
[4]
35. (a) s = 17, t = 90 (A1)(A1) 2
(b) (M1)(M2) 3
100 90 80 70 60 50 40 30 20 10
0 10 20 30 40 50 60 70 80 90 100
Marks
C
u
m
u
la
ti
v
e
fr
eq
u
en
cy
Award (M2) for 8, 9, 10 points plotted correctly, (M1) for 5, 6, 7 points plotted correctly, (M0) for 4 or less.
Accept a polygon or a curve.
(c) (i) Median mark = 48 (±1) (A1)
(ii) Lower quartile = 36 (±1) (A1)
(iii) Pass mark if 40% pass = 51 (±1) (M1)(A1) 4
Note: Follow through with candidate’s own graph.
Award (M0)(A1) ft if candidate correctly finds the grade (44) where 40% fail.
[9] 36. (a)
Time less than (mins) Cumulative frequency 10.5
15.5 20.5 25.5 30.5 35.5 40.5
0 7 20 45 73 93
100 (A2) 2
Note: Award (A1) for each correct column (b)
100 90 80 70 60 50 40 30 20 10
0 5 10 15 20 25 30 35 40
time
n
u
m
b
er
o
f
st
u
d
en
ts
c(i)
c(ii)
(A3) 3
Note: Award (A1) for the correct scale and labelling.
Award (A2) for plotting 6 or 7 points correctly, (A1) for plotting 4 or 5 points correctly.
(c) (i) 12 ± 1 students (allow ft) (A1)
(ii) 31 ± 0.5 minutes (allow ft) (A1) 2
[7] 37. (a)
L (cm) f Σf
≤ 29 2 2
≤ 31 4 6
≤ 33 8 14
≤ 35 21 35
≤ 37 30 65
≤ 39 18 83
≤ 41 12 95
≤ 43 5 100 (A2) 2
Notes: Award (Al) for four correct entries in the column headed Σf.
Award (A2) for all 8 correct. (b)
100 90 80 70 60 50 40 30 20 10 0
29 31 33 35 37 39 41 43 27
length (cm)L
cu m u la ti v e fr eq u en cy
(A3) 3
Notes: Award (Al) for both axes and correct scale.
Award [½ mark] for each correctly plotted point and round up to a maximum of [2 marks].
(c) (i) Median length of mackerel = 36 cm ± 0.2 cm (M1)
= 36 cm (A1)
(ii) Interquartile range of mackerel length = 3.8 ± 0.2 cm (M1)
= 4 cm (A1) 4*
*(read from candidate’s curve)
[9] 38. Unit penalty (UP) is applicable where indicated.
55 50 45 40 35 30 25 20 15 10 5 0 number of chickens 6 5 4 3 2 1 0 weight (kg)
(a) Dashed or solid line (using a ruler) passing through
mid-points of the 4 bars. (A1)
Starting and finishing at
0 , 2 1
and
0 , 2 1
5 respectively. (A1) (C2)
(b) 96 (A1) (C1)
UP (c) 3 ≤ weight < 4 kg. (Accept 3–4 kg) (A1) (C1) (d) For adding three heights or subtracting 14 from 96 (M1)
96 82
(0.854 or
48 41
, 85.4%) (ft)from (b). (A1)(ft) (C2)
[6]
39. (a) Discrete (A1) (C1)
2.73 (A1) (C2)
(c) 1.34 (A1) (C1)
Notes: for (b) and (c), if both mean and standard deviation given to 2 significant figures
Award (C1)(C0)(AP) for 2.7. Award (A1)(ft) for 1.3 ((AP) already deducted)
(d) Attempt to find their mean + their standard deviation (can be implied) (M1) 23, (ft)their mean and standard deviation. (A1)(ft) (C2)
[6]
40. (a) (i) 19.2 (G1)
(ii) 1.45 (G1)
(b) r = 0.942 (G1)
(c) Strong, positive correlation. (A1)(ft)(A1)(ft)
(d) (i) d = 11.5 (G1)
(ii) n = 11.5 × 19.6 – 100 (A1)(ft)
= 125 (accept 126)
Note: Answer must be a whole number
(e) It is unreliable to extrapolate outside the values given (outlier). (R1)
[8]
41. Unit penalty (UP) applies in part (c) in this question. (a)
(A1)(A1) (C2)
Note: (A1) for all correct heights, (A1) for all correct end points (39.5, 49.5 etc.).
Histogram must be drawn with a ruler (straight edge) and endpoints must be clear.
Award (A1) only if both correct histogram and correct frequency polygon drawn.
(b) 44.5 (A1) (C1)
(c) Mean =
42
... 18 5 . 54 6 5 .
44 × + × +
(M1)
Note: (M1) for a sum of frequencies multiplied by midpoint values divided by 42.
UP = 58.3 kg (A1)(ft) (C2)
Note: Award (A1)(A0)(AP) for 58.
(d) Standard deviation = 8.44 (A1) (C1)
Note: If (b)isgivenas45thenaward (b) 45 (A0)
(c) 58.8 kg (M1)(A1)(ft) or (C2)(ft) if no working seen. (d) 8.44 (C1)
[6] 42. (a) (i) Mean = (5.96 + 5.95 + 6.02 + 5.95 + 5.99) / 5 = 5.974 (5.97) (A1)
(ii) 100%
value actual
error error
% = ×
100% 5.974
5.974
6− ×
Note: (M1) for correctly substituted formula. Allow 0.503% as follow through from 5.97 Note: An answer of 0.433% is incorrect.
(b) number is 29.45728613
(i) Nearest integer = 29 (A1)
(ii) Standard form = 2.95 × 101 (accept 2.9 × 101) (A1)(ft)(A1) (C3)
Notes: Award (A1) for each correct term Award (A1)(A0) for 2.95 × 10
[6] 43. (a) 1.6 × 10–19, 9.8 × 10–18, π, 3.5, 0.006073 × 106, 60730, 6.073 × 105 (A4) (C4)
Notes: Award(A1)for π before 3.5
Award (A1) for 1.6 × 10–19 before 9.8 × 10–18
Award (A1) for the three numbers containing 6073 in the correct order. Award (A1) for the pair with negative indices placed before 3.5 and π and the remaining three numbers placed after (independently of the other three marks). Award (A3) for numbers given in correct decreasing order.
Award (A2) for decreasing order with at most 1 error
(b) The median is 3.5. (A1)(ft) (C1)
Note: Follow through from candidate’s list.
(c) π is irrational. (A1) (C1)
[6] 44. (a)
(A4)
Note: Award (A1) for correct scales and labels, (A3) for correct points, (A2) for 7 or 8 correct, (A1) for 5 or 6 correct.
(b) (i) 4 (G1)
(ii) 2.58 (G1)
(iii) 30 (G1)
(iv) 4.78 (G1)
Note: If wrong version of s.d. used in (ii), can (ft) in (iv) (5.07).
(c) L correctly plotted on graph and named (A1)(ft)
(d) (i) r = 0.986 (0.987) (G1)
(ii) (very) strong positive correlation (R1)(ft)(R1)(ft) (e) y = 1.83x + 22.7 (y = 1.825x + 22.7) (G1)(G1)
Award (G1) for y= 1.83x (1.825x), (G1) for 22.7
(f) Line drawn on graph. (A1)(A1)(ft)
Note: Award (A1) for passing through the mean point, (A1) for y intercept between 22 and 23.
Note: Allow margin of error of 0.2 from value on candidate’s diagram. (ii) Not possible to find an answer as the value lies too far outside the
given set of data. (R1)
[18]
45. (a) 80 × 62.1 + 2 ×11 = 4990 (M1)(A1)
Note: Award (M0)(A0) if 2 × 11 is subtracted and ft the remainder of the question to answers of 61.825 (or 61.8) and 59.7 respectively.
80 4990
= 62.375 (or 62.4 to 3 s.f.) (M1)(A1) (C4)
(b) 4990 + 56.8 × 60 = 8398 (M1)(A1)
140 8398
= 60.0 (3 s.f.) (M1)(A1) (C4)
Note: An answer of 60 (2 s.f.) with no working receives (G2) or with working using 4990 receives (M1)(A1)(M1)(A0) AP, however, if 80 × 62.4 is used then 60 is an exact answer and can receive all the marks.
[8] 46. (a) 35 . 5 4 . 12 5 . 44 −
= 6 (M1)(A1) (C2)
(b) 90 × 12.4 = 1116 (M1)(A1)
1116 – 44.5 – 43.2 = 1028.3 (M1)(A1)
88 3 . 1028
= 11.7 (M1)(A1) (C6)
Note: Award (M0)(A0) then ft for 88 × 12.4. Award (M0)(A0) for
90 1028.3
.
[8] 47. (a) (i) 1992 mean = $1.59, Sd = $0.727 (or 0.73) (A1) (A1)
(accept 0.777 or 0.78)
(ii) 2002 mean = $1.98, Sd = $0.635 (or 0.64) (A1) (A1) 4 (accept 0.679 or 0.68)
(b) (i) r =
= × = × = × 588 . 0 679 . 0 777 . 0 3104 . 0 or 672 . 0 635 . 0 727 . 0 3104 . 0 or 664 . 0 64 . 0 73 . 0 3104 . 0 (M1) (A1) OR
R = 0.672 (G2)
(ii) There is a weak positive correlation (R1) (R1) 4 (c) y – 1.98 = ( –1.59)
) 73 . 0 ( 3104 . 0
2 x (M1)
y = 0.582x + 1.05 (A1) (A1)
OR
y = 1.98 = ( –1.59) ) 727 . 0 ( 3104 . 0
2 x (M1)
y = 0.587x + 1.05 (A1) (A1)
OR
y = 0.588x + 1.05 (G3) 3
(d) y = 0.582 × 2.60 + 1.05
= $2.56 (A1)
OR
y = 0.587 × 2.60 + 1.05
= $2.58 (A1)
OR
= $2.58 (A1) 1 (e) Coffee – because it is the only item to go down in price. (A1) (R1)
OR
Rolls – because the price increased significantly. (A1) (R1) 2
[14] 48. (a)
x 155 161 173 150 182 165 170 185 175 145
y 50 75 80 46 81 79 64 92 74 108
100
90 80 70 60 50 40 30 20 10 0
20 40 60 80 100 120 140 160 180 200 weight
(kg)
Height (cm)
x y
(A2)(A2) 4
Notes: Award (A1) for axes correctly labelled, and (A1) for correct scales. Award (A1) for 4, 5 6, or 7 correctly plotted points, (A2) for 8 or more.
(b) Mean height = 166.1 = 166 (3 s.f.) (A1) 1
(c) Mean weight = 74.9 (3 s.f.) (A1) 1
(d) (i) Sx = 12.68 (A1)
Gradient =
2 2
) 68 . 12 (
31 . 44
= Sx Sxy
= 0.276 (M1)(AG)
(ii) y – 74.9 = 0.276(x – 166) (M1)
y = 0.276x + 29.1 (A1)
OR
y = 0.276x + 29.1 (G2)
(iii) Line on graph. (A2) 6
Note: Award (A1) for the y-intercept at 29.1, and (A1) for a straight line through (166, 74.9).
(e) (i) y = 0.276 × 190 + 29.1 (A1)
= 81.5 kg
(ii) 72 = 0.276x + 29.1 x =
276 . 0
1 . 29 72−
= 155 cm. (A1)
OR
From the graph (A1)
(i) y = 81 (±1) (A1)
Note: Follow through with candidate’s line.
(f) The “line of best fit” becomes closer to the remaining points. (R1)
OR
Gradient becomes steeper and the line is more accurate ‘best fit’. (R1)
OR
Any reasonable explanation. (Line becomes y = 1.10x – 113) (R1) 1
[15] 49. (a) (i)
5
190 178 162 173
182+ + + +
= 177 cm (A1)
(ii)
5
75 66 60 68
73+ + + +
= 68.4 kg (A1)
(b)
190 185 180 175 170 165 160
0
60 65 70 75
weight (kg)
height (cm)
M
(A1)(A1)
Note: Award (A1) for at least 3 points plotted correctly; (A1) for a line of best fit through (68.4, 177)
