Sec 4 E-Math Statistics worksheet with answers

A free Sec 4 E-Math worksheet on statistics: 9 questions, 30 marks, short questions first and then structured ones, in the school exam style. The answers and full worked solutions are on the last page.

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Sec 4 E-Math worksheet: Statistics

Total: 30 marksSuggested time: 45 minutes

Name: ______________________________Class: ____________ Date: ____________

Answer all questions. Show all your working. Give non-exact answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless the question says otherwise.

Section A

The box-and-whisker plot shows the times, in minutes, that

120

students took to finish a task. Estimate the number of students who took between

26

and

40

minutes.

Answer[1]

The cumulative frequency curve shows the marks of

50

students. Use the curve to estimate the percentage of the students who scored

25

marks or fewer.

Answer%[1]

The cumulative frequency curve shows the marks of

80

students in a test. Use the curve to estimate the number of students who scored at most

45

marks.

Answer[1]

Marks, $x$	Frequency
$0 < x \le 10$	2
$10 < x \le 20$	5
$20 < x \le 30$	9
$30 < x \le 40$	14
$40 < x \le 50$	7
$50 < x \le 60$	3

The table shows the marks of

40

students. Using the mid-values of the classes, estimate the mean mark.

Answer[2]

(a)In a test, class

A

had a mean of

62

marks and a standard deviation of

6.3

marks. Class

B

had a mean of

65

marks and a standard deviation of

7.9

marks. Which class did better on average?

Answer (a)[1]

(b)The numbers of books read by some students in two classes are shown. Class

A

2, \ 12, \ 7, \ 4, \ 10

. Class

B

8, \ 20, \ 11, \ 18, \ 13

. Which class's numbers of books read were more consistent?

Answer (b)[1]

Find the median of

12, \ 7, \ 15, \ 9, \ 21, \ 18, \ 11

Answer[2]

Section B

The table shows the distance,

d

km, that some workers travel to work. An estimate of the mean is

23

km.

Distance ( $d$ km)	Frequency
$0 < d \le 10$	6
$10 < d \le 20$	13
$20 < d \le 30$	$k$
$30 < d \le 50$	4
$50 < d \le 80$	3

(a)Find the value of

k

Answer (a)[3]

(b)Calculate an estimate of the standard deviation. Give your answer correct to 3 significant figures.

Answer (b)km[2]

(c)Which class contains the median?

Answer (c)[1]

(d)One of the workers is chosen at random. Find the probability that this person travels more than

50

km. Give your answer as a fraction in its simplest form.

Answer (d)[1]

The table shows the time taken,

t

minutes, by each of the

37

students in Class

P

to travel to school.

Time ( $t$ minutes)	Frequency
$0 < t \le 10$	8
$10 < t \le 20$	9
$20 < t \le 30$	13
$30 < t \le 40$	$p$
$40 < t \le 50$	3

(a)Find the value of

p

Answer (a)[1]

(b)Calculate an estimate of the mean time. Give your answer correct to 3 significant figures.

Answer (b)min[2]

(c)Calculate an estimate of the standard deviation. Give your answer correct to 3 significant figures.

Answer (c)min[2]

(d)For Class

Q

, the mean is

17.3

minutes and the standard deviation is

15.1

minutes. Which group took less time, on average, to travel to school?

Answer (d)[1]

(e)Which group's times were more consistent?

Answer (e)[1]

The seven numbers below are written in order of size, from the smallest.

7,\ f,\ 21,\ g,\ 26,\ 31,\ h

. The range of the numbers is

25

, the median is

25

and the interquartile range is

17

(a)Find the value of

h

Answer (a)[1]

(b)Find the value of

g

Answer (b)[1]

(c)Find the value of

f

Answer (c)[2]

(d)Find the mean of the seven numbers, correct to 3 significant figures.

Answer (d)[1]

(e)

5

is added to each of the seven numbers. Write down the new mean, correct to 3 significant figures.

Answer (e)[1]

(f)How does the standard deviation change when

5

is added to each number?

Answer (f)[1]

END OF PAPER

Answers and worked solutions

Question 1

Answer:

\dfrac{1}{2} \times 120 = 60

[1]

Half of the students.

\dfrac{1}{2} \times 120 = 60

Question 2

Answer:

\dfrac{11.4}{50} \times 100\% \approx 22.8\%

[1]

About

11.4

students scored

25

marks or fewer.

\dfrac{11.4}{50} \times 100\% \approx 22.8\%

Question 3

Answer: The curve gives about

68

students with at most

45

marks.[1]

The curve gives about

68

students with at most

45

marks.

Question 4

Answer:

\text{Mean} = 32

[2]

x = 5, 15, 25, 35, 45, 55

\Sigma f = 40

\Sigma fx = 1280

\mathstrut \text{Mean} = \dfrac{1280}{40}

\text{Mean} = 32

Question 5

(a)Answer: B[1]

Better on average means the higher mean.Class A's is lower.Class B.

(b)Answer: A[1]

More consistent means less spread: the smaller standard deviation.

A: 3.69

B: 4.43

Class A's is smaller.Class A.

Question 6

Answer:

\text{Median} = 12

[2]

7, \ 9, \ 11, \ 12, \ 15, \ 18, \ 21

n = 7

The middle value is the 4th.

\text{Median} = 12

Question 7

(a)Answer:

k = 9

[3]

\text{Mid-values: } 5,\ 15,\ 25,\ 40,\ 65

\dfrac{580 + 25k}{26 + k} = 23

580 + 25k = 598 + 23k

2k = 18

k = 9

(b)Answer:

16.3

km[2]

\Sigma f = 35

\Sigma fx = 805

\Sigma fx^2 = 27775

\text{sd} = \sqrt{\dfrac{27775}{35} - \left(\dfrac{805}{35}\right)^2}

= 16.3

(c)Answer:

10 < d \le 20

[1]

The median is the

18

th of the

35

values.Running totals: 6, 19, 28, 32, 35: the median is in

10 < d \le 20

(d)Answer:

\dfrac{3}{35}

[1]

P = \dfrac{3}{35}

\phantom{P} = \dfrac{3}{35}

Question 8

(a)Answer:

p = 4

[1]

p = 37 - 33

p = 4

(b)Answer:

20.9

min[2]

\text{Mid-values: } 5, \ 15, \ 25, \ 35, \ 45

\Sigma fx = 8 \times 5 + 9 \times 15 + 13 \times 25 + 4 \times 35 + 3 \times 45

\phantom{\Sigma fx} = 775

\text{Mean} = \dfrac{775}{37}

\phantom{\text{Mean}} = 20.9

(c)Answer:

11.7

min[2]

\Sigma fx^2 = 8 \times 5^2 + 9 \times 15^2 + 13 \times 25^2 + 4 \times 35^2 + 3 \times 45^2

\phantom{\Sigma fx^2} = 21325

\text{sd} = \sqrt{\dfrac{21325}{37} - \left(\dfrac{775}{37}\right)^2}

\phantom{\text{sd}} = 11.7

(d)Answer: Class

Q

: its mean is smaller[1]

20.9 > 17.3

, so Class

Q

took less time, on average, to travel to school.

(e)Answer: Class

P

: its standard deviation is smaller[1]

11.7 < 15.1

, so the times of Class

P

were more consistent.

Question 9

(a)Answer:

h = 32

[1]

\text{Range} = h - 7

\phantom{\text{Range}} = 25

h = 32

(b)Answer:

g = 25

[1]

\text{The median is the 4th value: } g = 25

(c)Answer:

f = 14

[2]

\text{Lower half: } 7,\ f,\ 21

, so

Q_1 = f

\text{Upper half: } 26,\ 31,\ h

, so

Q_3 = 31

31 - f = 17

f = 14

(d)Answer:

22.3

[1]

\text{Mean} = \dfrac{7 + 14 + 21 + 25 + 26 + 31 + 32}{7}

= \dfrac{156}{7}

= 22.3

(e)Answer:

27.3

[1]

Every value goes up by

5

, so the mean goes up by

5

\text{New mean} = 27.3

(f)Answer: It stays the same: the numbers are just as spread out.[1]

Every value moves by the same amount, so the distances from the mean, and the standard deviation, do not change.

Practice questions by skill

Sec 4 E-Math Statistics worksheet with answers

In the print window, choose Save as PDF to download it.

Need the formulas first? Read the Statistics notes. Or practise it online with step-by-step help when you get one wrong.

Marks,

x

Frequency

0 < x \le 10

10 < x \le 20

20 < x \le 30

30 < x \le 40

40 < x \le 50

50 < x \le 60

Section B

The table shows the distance,

d

km, that some workers travel to work. An estimate of the mean is

23

km.

Distance ( $d$ km)	Frequency
$0 < d \le 10$	6
$10 < d \le 20$	13
$20 < d \le 30$	$k$
$30 < d \le 50$	4
$50 < d \le 80$	3

(a)Find the value of

k

Answer (a)[3]

(b)Calculate an estimate of the standard deviation. Give your answer correct to 3 significant figures.

Answer (b)km[2]

(c)Which class contains the median?

Answer (c)[1]

(d)One of the workers is chosen at random. Find the probability that this person travels more than

50

km. Give your answer as a fraction in its simplest form.

Answer (d)[1]

Time (

t

minutes)

Frequency

0 < t \le 10

10 < t \le 20

20 < t \le 30

30 < t \le 40

p

40 < t \le 50

Answers and worked solutions

Question 1

Answer:

\dfrac{1}{2} \times 120 = 60

[1]

Half of the students.

\dfrac{1}{2} \times 120 = 60

Question 2

Answer:

\dfrac{11.4}{50} \times 100\% \approx 22.8\%

[1]

About

11.4

students scored

25

marks or fewer.

\dfrac{11.4}{50} \times 100\% \approx 22.8\%

Question 3

Answer: The curve gives about

68

students with at most

45

marks.[1]

The curve gives about

68

students with at most

45

marks.

Question 4

Answer:

\text{Mean} = 32

[2]

x = 5, 15, 25, 35, 45, 55

\Sigma f = 40

\Sigma fx = 1280

\mathstrut \text{Mean} = \dfrac{1280}{40}

\text{Mean} = 32

Question 5

(a)Answer: B[1]

Better on average means the higher mean.Class A's is lower.Class B.

(b)Answer: A[1]

More consistent means less spread: the smaller standard deviation.

A: 3.69

B: 4.43

Class A's is smaller.Class A.

Question 6

Answer:

\text{Median} = 12

[2]

7, \ 9, \ 11, \ 12, \ 15, \ 18, \ 21

n = 7

The middle value is the 4th.

\text{Median} = 12

Question 7

(a)Answer:

k = 9

[3]

\text{Mid-values: } 5,\ 15,\ 25,\ 40,\ 65

\dfrac{580 + 25k}{26 + k} = 23

580 + 25k = 598 + 23k

2k = 18

k = 9

(b)Answer:

16.3

km[2]

\Sigma f = 35

\Sigma fx = 805

\Sigma fx^2 = 27775

\text{sd} = \sqrt{\dfrac{27775}{35} - \left(\dfrac{805}{35}\right)^2}

= 16.3

(c)Answer:

10 < d \le 20

[1]

The median is the

18

th of the

35

values.Running totals: 6, 19, 28, 32, 35: the median is in

10 < d \le 20

(d)Answer:

\dfrac{3}{35}

[1]

P = \dfrac{3}{35}

\phantom{P} = \dfrac{3}{35}

Question 8

(a)Answer:

p = 4

[1]

p = 37 - 33

p = 4

(b)Answer:

20.9

min[2]

\text{Mid-values: } 5, \ 15, \ 25, \ 35, \ 45

\Sigma fx = 8 \times 5 + 9 \times 15 + 13 \times 25 + 4 \times 35 + 3 \times 45

\phantom{\Sigma fx} = 775

\text{Mean} = \dfrac{775}{37}

\phantom{\text{Mean}} = 20.9

(c)Answer:

11.7

min[2]

\Sigma fx^2 = 8 \times 5^2 + 9 \times 15^2 + 13 \times 25^2 + 4 \times 35^2 + 3 \times 45^2

\phantom{\Sigma fx^2} = 21325

\text{sd} = \sqrt{\dfrac{21325}{37} - \left(\dfrac{775}{37}\right)^2}

\phantom{\text{sd}} = 11.7

(d)Answer: Class

Q

: its mean is smaller[1]

20.9 > 17.3

, so Class

Q

took less time, on average, to travel to school.

(e)Answer: Class

P

: its standard deviation is smaller[1]

11.7 < 15.1

, so the times of Class

P

were more consistent.

Question 9

(a)Answer:

h = 32

[1]

\text{Range} = h - 7

\phantom{\text{Range}} = 25

h = 32

(b)Answer:

g = 25

[1]

\text{The median is the 4th value: } g = 25

(c)Answer:

f = 14

[2]

\text{Lower half: } 7,\ f,\ 21

, so

Q_1 = f

\text{Upper half: } 26,\ 31,\ h

, so

Q_3 = 31

31 - f = 17

f = 14

(d)Answer:

22.3

[1]

\text{Mean} = \dfrac{7 + 14 + 21 + 25 + 26 + 31 + 32}{7}

= \dfrac{156}{7}

= 22.3

(e)Answer:

27.3

[1]

Every value goes up by

5

, so the mean goes up by

5

\text{New mean} = 27.3

(f)Answer: It stays the same: the numbers are just as spread out.[1]

Every value moves by the same amount, so the distances from the mean, and the standard deviation, do not change.