Sec 4 G2 Maths Statistics worksheet with answers

A free Sec 4 G2 Maths worksheet on statistics: 7 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 G2 Maths worksheet: Statistics

Total: 30 marksSuggested time: 50 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 cumulative frequency curve shows the marks of

50

students. Use the curve to estimate the median mark.

Answer[1]

The cumulative frequency curve shows the marks of

50

students in a test. Use the curve to estimate the number of students who scored more than

15

marks.

Answer[2]

(a)In a test, class

A

had a mean of

71

marks and a standard deviation of

8.4

marks. Class

B

had a mean of

68

marks and a standard deviation of

12.1

marks. Which class did better on average?

Answer (a)[1]

(b)In a test, class

A

had a mean of

55

marks and a standard deviation of

11.5

marks. Class

B

had a mean of

58

marks and a standard deviation of

10.2

marks. Which class's results were more consistent?

Answer (b)[1]

(a)Find the standard deviation of

30, \ 34, \ 35, \ 39, \ 42

, correct to 3 significant figures.

Answer (a)[3]

(b)For a set of

8

values,

\Sigma x = 48

and

\Sigma x^2 = 424

. Find the standard deviation, correct to 3 significant figures.

Answer (b)[2]

Section B

The table shows the times,

t

minutes, that

40

customers waited at a counter. An estimate of the mean waiting time is

40

minutes.

Time ( $t$ minutes)	Frequency
$0 < t \le 20$	$p$
$20 < t \le 40$	$q$
$40 < t \le 60$	11
$60 < t \le 80$	3
$80 < t \le 100$	5

(a)Show that

p + q = 21

[1]

(b)Show that

p + 3q = 39

[2]

(c)Solve the two equations to find

p

and

q

. Give your answer as

p

q

Answer (c)[2]

(d)Which class contains the median waiting time?

Answer (d)[1]

The table shows the time,

t

minutes, that some customers waited at a clinic. An estimate of the mean is

13.75

minutes.

Time ( $t$ minutes)	Frequency
$0 < t \le 5$	8
$5 < t \le 10$	$k$
$10 < t \le 20$	11
$20 < t \le 30$	6
$30 < t \le 50$	2

(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)min[2]

(c)Which class contains the median?

Answer (c)[1]

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

30

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

Answer (d)[1]

The stem-and-leaf diagram shows the marks scored by the

20

students in Class

A

in a mathematics test.

(a)Find the median mark.

Answer (a)[1]

(b)Find the interquartile range.

Answer (b)[2]

(c)The pass mark is

50

. Find the percentage of the students who passed.

Answer (c)%[1]

(d)For Class

B

, the median is

50.5

marks and the interquartile range is

12.5

marks. Which statement is correct?

Answer (d)[1]

(e)Two of the students in Class

A

are chosen at random. Find the probability that both scored at least

50

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

Answer (e)[2]

END OF PAPER

Answers and worked solutions

Question 1

Answer:

\text{Median} \approx 30.0

[1]

\mathstrut \dfrac{50}{2} = 25

\text{Median} \approx 30.0

Question 2

Answer:

\text{Number} \approx 47

[2]

The curve gives about

3

students with at most

15

marks.

\mathstrut 50 - 3

\text{Number} \approx 47

Question 3

(a)Answer: A[1]

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

(b)Answer: B[1]

More consistent means less spread: the smaller standard deviation.Class A's is larger.Class B.

Question 4

(a)Answer:

\phantom{\text{sd}} = 4.15

(3 s.f.)[3]

\text{sd} = \sqrt{\dfrac{\Sigma x^2}{n} - \left(\dfrac{\Sigma x}{n}\right)^2}

n = 5

\mathstrut \Sigma x = 30 + 34 + 35 + 39 + 42

\Sigma x = 180

\mathstrut \Sigma x^2 = 30^2 + 34^2 + 35^2 + 39^2 + 42^2

\Sigma x^2 = 6566

\mathstrut \text{sd} = \sqrt{\dfrac{6566}{5} - \left(\dfrac{180}{5}\right)^2}

\text{sd} = 4.1472\ldots

\phantom{\text{sd}} = 4.15

(3 s.f.)

(b)Answer:

\phantom{\text{sd}} = 4.12

(3 s.f.)[2]

\text{sd} = \sqrt{\dfrac{\Sigma x^2}{n} - \left(\dfrac{\Sigma x}{n}\right)^2}

\mathstrut \text{sd} = \sqrt{\dfrac{424}{8} - \left(\dfrac{48}{8}\right)^2}

\text{sd} = 4.1231\ldots

\phantom{\text{sd}} = 4.12

(3 s.f.)

Question 5

(a)[1]

11 + 3 + 5 + p + q = 40

p + q = 21

(b)[2]

\text{Mid-values: } 10,\ 30,\ 50,\ 70,\ 90

\dfrac{1210 + 10p + 30q}{40} = 40

10p + 30q = 390

p + 3q = 39

(c)Answer:

p = 12

q = 9

[2]

p + q = 21

... (1)

p + 3q = 39

... (2)

p = 12

q = 9

(d)Answer:

20 < t \le 40

[1]

Running totals: 12, 21, 32, 35, 40. The median is the mean of the

20

th and

21

th values: in

20 < t \le 40

Question 6

(a)Answer:

k = 7

[3]

\text{Mid-values: } 2.5,\ 7.5,\ 15,\ 25,\ 40

\dfrac{415 + 7.5k}{27 + k} = 13.75

415 + 7.5k = 371.25 + 13.75k

-6.25k = -43.75

k = 7

(b)Answer:

10.1

minutes[2]

\Sigma f = 34

\Sigma fx = 467.5

\Sigma fx^2 = 9868.75

\text{sd} = \sqrt{\dfrac{9868.75}{34} - \left(\dfrac{467.5}{34}\right)^2}

= 10.1

(c)Answer:

10 < t \le 20

[1]

The median is halfway between the

17

th and

18

th of the

34

values.Running totals: 8, 15, 26, 32, 34: the median is in

10 < t \le 20

(d)Answer:

\dfrac{1}{17}

[1]

P = \dfrac{2}{34}

\phantom{P} = \dfrac{1}{17}

Question 7

(a)Answer:

55.5

[1]

\text{Median} = \dfrac{55 + 56}{2}

\phantom{\text{Median}} = 55.5

(the

10

th and

11

th values)

(b)Answer:

17.5

[2]

Q_1 = \dfrac{50 + 51}{2}

\phantom{Q_1} = 50.5

(median of the lower

10

values)

Q_3 = \dfrac{67 + 69}{2}

\phantom{Q_3} = 68

(median of the upper

10

values)

\text{IQR} = 68 - 50.5

\phantom{\text{IQR}} = 17.5

(c)Answer:

80\%

[1]

16

of the

20

scored

50

or more

\dfrac{16}{20} \times 100\% = 80\%

(d)Answer: Class

A

did better on average, and Class

B

was more consistent.[1]

\text{Medians: } 55.5 > 50.5

, so Class

A

did better on average.

\text{IQRs: } 17.5 > 12.5

, so Class

B

was more consistent.

(e)Answer:

\dfrac{12}{19}

[2]

P = \dfrac{16}{20} \times \dfrac{15}{19}

\phantom{P} = \dfrac{12}{19}

Sec 4 G2 Maths 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.

Section B

The table shows the times,

t

minutes, that

40

customers waited at a counter. An estimate of the mean waiting time is

40

minutes.

Time ( $t$ minutes)	Frequency
$0 < t \le 20$	$p$
$20 < t \le 40$	$q$
$40 < t \le 60$	11
$60 < t \le 80$	3
$80 < t \le 100$	5

(a)Show that

p + q = 21

[1]

(b)Show that

p + 3q = 39

[2]

(c)Solve the two equations to find

p

and

q

. Give your answer as

p

q

Answer (c)[2]

(d)Which class contains the median waiting time?

Answer (d)[1]

Time (

t

minutes)

Frequency

0 < t \le 5

5 < t \le 10

k

10 < t \le 20

20 < t \le 30

30 < t \le 50

Answers and worked solutions

Question 1

Answer:

\text{Median} \approx 30.0

[1]

\mathstrut \dfrac{50}{2} = 25

\text{Median} \approx 30.0

Question 2

Answer:

\text{Number} \approx 47

[2]

The curve gives about

3

students with at most

15

marks.

\mathstrut 50 - 3

\text{Number} \approx 47

Question 3

(a)Answer: A[1]

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

(b)Answer: B[1]

More consistent means less spread: the smaller standard deviation.Class A's is larger.Class B.

Question 4

(a)Answer:

\phantom{\text{sd}} = 4.15

(3 s.f.)[3]

\text{sd} = \sqrt{\dfrac{\Sigma x^2}{n} - \left(\dfrac{\Sigma x}{n}\right)^2}

n = 5

\mathstrut \Sigma x = 30 + 34 + 35 + 39 + 42

\Sigma x = 180

\mathstrut \Sigma x^2 = 30^2 + 34^2 + 35^2 + 39^2 + 42^2

\Sigma x^2 = 6566

\mathstrut \text{sd} = \sqrt{\dfrac{6566}{5} - \left(\dfrac{180}{5}\right)^2}

\text{sd} = 4.1472\ldots

\phantom{\text{sd}} = 4.15

(3 s.f.)

(b)Answer:

\phantom{\text{sd}} = 4.12

(3 s.f.)[2]

\text{sd} = \sqrt{\dfrac{\Sigma x^2}{n} - \left(\dfrac{\Sigma x}{n}\right)^2}

\mathstrut \text{sd} = \sqrt{\dfrac{424}{8} - \left(\dfrac{48}{8}\right)^2}

\text{sd} = 4.1231\ldots

\phantom{\text{sd}} = 4.12

(3 s.f.)

Question 5

(a)[1]

11 + 3 + 5 + p + q = 40

p + q = 21

(b)[2]

\text{Mid-values: } 10,\ 30,\ 50,\ 70,\ 90

\dfrac{1210 + 10p + 30q}{40} = 40

10p + 30q = 390

p + 3q = 39

(c)Answer:

p = 12

q = 9

[2]

p + q = 21

... (1)

p + 3q = 39

... (2)

p = 12

q = 9

(d)Answer:

20 < t \le 40

[1]

Running totals: 12, 21, 32, 35, 40. The median is the mean of the

20

th and

21

th values: in

20 < t \le 40

Question 6

(a)Answer:

k = 7

[3]

\text{Mid-values: } 2.5,\ 7.5,\ 15,\ 25,\ 40

\dfrac{415 + 7.5k}{27 + k} = 13.75

415 + 7.5k = 371.25 + 13.75k

-6.25k = -43.75

k = 7

(b)Answer:

10.1

minutes[2]

\Sigma f = 34

\Sigma fx = 467.5

\Sigma fx^2 = 9868.75

\text{sd} = \sqrt{\dfrac{9868.75}{34} - \left(\dfrac{467.5}{34}\right)^2}

= 10.1

(c)Answer:

10 < t \le 20

[1]

The median is halfway between the

17

th and

18

th of the

34

values.Running totals: 8, 15, 26, 32, 34: the median is in

10 < t \le 20

(d)Answer:

\dfrac{1}{17}

[1]

P = \dfrac{2}{34}

\phantom{P} = \dfrac{1}{17}

Question 7

(a)Answer:

55.5

[1]

\text{Median} = \dfrac{55 + 56}{2}

\phantom{\text{Median}} = 55.5

(the

10

th and

11

th values)

(b)Answer:

17.5

[2]

Q_1 = \dfrac{50 + 51}{2}

\phantom{Q_1} = 50.5

(median of the lower

10

values)

Q_3 = \dfrac{67 + 69}{2}

\phantom{Q_3} = 68

(median of the upper

10

values)

\text{IQR} = 68 - 50.5

\phantom{\text{IQR}} = 17.5

(c)Answer:

80\%

[1]

16

of the

20

scored

50

or more

\dfrac{16}{20} \times 100\% = 80\%

(d)Answer: Class

A

did better on average, and Class

B

was more consistent.[1]

\text{Medians: } 55.5 > 50.5

, so Class

A

did better on average.

\text{IQRs: } 17.5 > 12.5

, so Class

B

was more consistent.

(e)Answer:

\dfrac{12}{19}

[2]

P = \dfrac{16}{20} \times \dfrac{15}{19}

\phantom{P} = \dfrac{12}{19}