Standard Deviation - Worksheets, Questions and Revision

15 original exam-style questions - 11 pages of questions with a full mark scheme - free printable PDF.

Download PDF Jump to mark scheme (page 12)Read the revision guide

Next in this strand

Revision Library

revisionlibrary.co.uk

HIGHER

H04 Standard Deviation

EDEXCEL 1ST0 · Calculator allowed · about 90 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

A cafe manager records how long, in minutes, 6 customers wait to be served:

3, 5, 4, 6, 3, 9

(a)Work out the mean waiting time.(2)

(b)Work out the range of the waiting times.(1)

(Total for Question 1 is 3 marks)

For each statement about standard deviation, write down whether it is True or False.

(i)The standard deviation of a data set can never be negative.(1)

(ii)If every value in a data set is identical, the standard deviation is 0.(1)

(iii)A data set with a larger standard deviation always has a larger mean than a data set with a smaller standard deviation.(1)

(iv)The standard deviation of a data set is measured in the same units as the original data.(1)

(Total for Question 2 is 4 marks)

A hockey team scores the following number of goals in 5 matches:

2, 5, 3, 4, 6

The mean number of goals scored is 4.

Key formula: standard deviation = √sum of (x - mean)² / n

(a)Show that the sum of (x - mean)² for this data set is 10.(2)

(b)Hence work out the standard deviation of the number of goals scored, giving your answer correct to 3 significant figures.(2)

(Total for Question 3 is 4 marks)

5 friends compare how many text messages they sent yesterday:

8, 12, 10, 14, 6

An equivalent formula for standard deviation is:

standard deviation = √sum of x² / n - mean²

(a)Show that the sum of x² for this data set is 540.(2)

(b)Write down the mean number of texts sent.(1)

(c)Using the formula above, calculate the standard deviation of the number of texts sent, giving your answer correct to 3 significant figures.(2)

(Total for Question 4 is 5 marks)

Consider the evenly-spaced sequence of numbers:

5, 10, 15, 20, 25

(a)Using the symmetry of the sequence, write down its mean.(1)

(b)Calculate the standard deviation of the sequence, giving your answer correct to 3 significant figures.(3)

(Total for Question 5 is 4 marks)

The resistance, in ohms, of 5 electrical components is measured:

18, 22, 19, 21, 20

(a)Write down the formula that connects the variance of a data set to its standard deviation.(1)

(b)Calculate the variance of the 5 resistance readings.(2)

(c)Hence write down the standard deviation of the resistance readings, correct to 3 significant figures, stating its units.(2)

(d)Give one advantage of using the standard deviation, rather than the variance, to describe the spread of the resistance readings.(1)

(Total for Question 6 is 6 marks)

Two market stalls record their sales, in pounds, over the same 5 days.

Day	1	2	3	4	5
Stall A sales (£)	40	45	38	42	45
Stall B sales (£)	20	60	15	70	45

(a)Show that the mean daily sales for both Stall A and Stall B is £42.(2)

(b)Calculate the standard deviation of Stall A's daily sales, giving your answer correct to 3 significant figures.(3)

(c)Calculate the standard deviation of Stall B's daily sales, giving your answer correct to 3 significant figures.(3)

(d)Using the standard deviations, write down which stall has more consistent daily sales, giving a reason.(2)

(Total for Question 7 is 10 marks)

The bar chart shows the number of siblings for each of 20 pupils in a class survey.

Number of siblings	0	1	2	3	4
Frequency	3	7	6	3	1

Key formula for frequency data: standard deviation = √sum of fx² / sum of f - mean²

(a)Write down the modal number of siblings.(1)

(b)Show that the sum of fx for this data is 32.(2)

(c)Hence write down the mean number of siblings for this class.(1)

(d)Show that the sum of fx² for this data is 74.(2)

(e)Using the formula above, calculate the standard deviation of the number of siblings, giving your answer correct to 3 significant figures.(3)

(Total for Question 8 is 9 marks)

A gardener records the height, h cm, of 30 plants in a grouped frequency table.

Height, h (cm)	0 < h ≤ 10	10 < h ≤ 20	20 < h ≤ 30	30 < h ≤ 40
Frequency	4	9	11	6

(a)Write down the midpoint of the class 20 < h ≤ 30.(1)

(b)Using midpoints 5, 15, 25 and 35, show that an estimate for the sum of fx is 640.(2)

(c)Hence write down an estimate for the mean height of the plants, correct to 3 significant figures.(1)

(d)Show that an estimate for the sum of fx² is 16350.(2)

(e)Using standard deviation = √sum of fx² / sum of f - mean², calculate an estimate for the standard deviation of the plant heights, giving your answer correct to 3 significant figures.(3)

(Total for Question 9 is 9 marks)

The masses, in grams, of 5 packages are:

1024, 1031, 1019, 1028, 1022

The masses, x, are coded using y = x - 1020.

(a)Explain why coding the data using y = x - 1020 makes the values easier to work with.(1)

(b)Show that the mean of the coded data, y, is 4.8.(2)

(c)Hence write down the mean mass, x, of the original packages.(1)

(d)Calculate the standard deviation of the coded data, y, giving your answer correct to 3 significant figures.(3)

(e)Write down the standard deviation of the original masses, x, giving a reason for your answer.(1)

(Total for Question 10 is 8 marks)

A bakery manager wants to compare the consistency of loaf weights, in grams, produced by two ovens.

(a)Plan: Give one reason why the manager should weigh a sample of several loaves from each oven, rather than comparing just one loaf from each.(1)

(b)Collect: Give one step the manager should take when collecting the two samples to make the comparison fair.(1)

(c)Process: She weighs 5 loaves from each oven (grams):

Oven A: 798, 802, 805, 797, 803
Oven B: 780, 815, 790, 810, 805

Show that the mean loaf weight is 801 g for Oven A and 800 g for Oven B, and calculate the standard deviation of each sample, giving your answers correct to 3 significant figures.(8)

(d)Interpret: Using the standard deviations, write a conclusion comparing the consistency of loaf weights produced by the two ovens.(2)

(Total for Question 11 is 12 marks)

An athlete's first 4 javelin throws, in metres, are:

42, 45, 40, 41

(a)Find the mean of the original 4 throws.(1)

(b)Show that the standard deviation of the original 4 throws is 1.87 m, correct to 3 significant figures.(3)

(c)A fifth throw of 42 m is added to the data set. Work out the new standard deviation of all 5 throws, giving your answer correct to 3 significant figures.(3)

(d)Explain why adding this fifth throw decreased the standard deviation, even though it did not change the mean.(2)

(Total for Question 12 is 9 marks)

A student is asked to find the standard deviation of the data set 4, 8, 6, 10, 2, which has a mean of 6. Here is the student's working:

sum of (x - mean)² = 40
standard deviation = √sum of (x - mean)² = √40 = 6.32 (3 sf)

(a)Identify the error in the student's working.(1)

(b)Work out the correct value of the standard deviation, giving your answer correct to 3 significant figures.(2)

(Total for Question 13 is 3 marks)

A survey records 8 commute times, in minutes:

22, 25, 24, 23, 45, 26, 24, 23

(a)Write down the value in the sample that looks unusually different from the rest.(1)

(b)Show that the mean commute time is 26.5 minutes.(1)

(c)Show that the standard deviation of the commute times is 7.09 minutes, correct to 3 significant figures.(3)

(d)One rule for identifying an outlier is a value that is more than 2 standard deviations from the mean. Work out the boundary values for outliers using this rule.(2)

(e)Hence state which commute time in the sample should be classed as an outlier, giving a reason.(1)

(Total for Question 14 is 8 marks)

5 students' exam scores, out of 200, are:

120, 140, 160, 100, 180

The scores, x, are coded using y = (x - 100)/20.

(a)Using the coding y = (x - 100)/20, show that the coded values are 1, 2, 3, 0 and 4.(2)

(b)Write down the mean of the coded data, y.(1)

(c)Calculate the standard deviation of the coded data, y, giving your answer correct to 3 significant figures.(3)

(d)Hence write down the mean and the standard deviation of the original exam scores, x.(2)

(Total for Question 15 is 8 marks)

Mark scheme · H04 Standard Deviation

Question 1

(a) M1 (3+5+4+6+3+9)/6, oe (adds all six values and divides by 6)
(a) A1 5 (minutes) cao
(a) Answer: 5 minutes
(b) B1 6 (minutes) cao
(b) Answer: 6 minutes

Question 2

(i) B1 True cao
(i) Answer: True
(ii) B1 True cao
(ii) Answer: True
(iii) B1 False cao
(iii) Answer: False
(iv) B1 True cao
(iv) Answer: True

Question 3

(a) M1 deviations -2, 1, -1, 0, 2 (or squares 4, 1, 1, 0, 4), oe
(a) A1 10 cso
(a) Answer: 10
(b) M1 10/5 (=2), oe
(b) A1 1.41 (awrt)
(b) Answer: 1.41 (goals)

Question 4

(a) M1 64, 144, 100, 196, 36, oe (correct squares of each value)
(a) A1 540 cso
(a) Answer: 540
(b) B1 10 cao
(b) Answer: 10
(c) M1 540/5 - 10² (=108-100=8), oe
(c) A1 2.83 (awrt)
(c) Answer: 2.83 (texts)

Question 5

(a) B1 15 cao
(a) Answer: 15
(b) M1 deviations -10, -5, 0, 5, 10, so sum of (x-mean)² = 250, oe
(b) M1 250/5 (=50)
(b) A1 7.07 (awrt)
(b) Answer: 7.07

Question 6

(a) B1 variance = (standard deviation)², oe (e.g. standard deviation = √variance)
(a) Answer: variance = (standard deviation)²
(b) M1 deviations -2, 2, -1, 1, 0, so sum of (x-mean)² = 10, oe
(b) A1 variance = 2 cao
(b) Answer: 2 (ohms squared)
(c) M1 √2, ft their variance from (b)
(c) A1 1.41 ohms (awrt), with correct unit (not ohms squared)
(c) Answer: 1.41 ohms
(d) B1 the standard deviation is in the same units as the original data (ohms), so it is easier to interpret and compare directly with the data values, unlike the variance, which is in squared units, oe
(d) Answer: The standard deviation is in the same units as the data (ohms), making it easier to interpret than the variance (ohms squared).

Question 7

(a) B1 Stall A: (40+45+38+42+45)/5 = 210/5 = 42
(a) B1 Stall B: (20+60+15+70+45)/5 = 210/5 = 42
(a) Answer: £42 for both stalls
(b) M1 deviations -2, 3, -4, 0, 3, so sum of (x-mean)² = 38, oe
(b) M1 38/5 (=7.6)
(b) A1 £2.76 (awrt)
(b) Answer: £2.76
(c) M1 deviations -22, 18, -27, 28, 3, so sum of (x-mean)² = 2330, oe
(c) M1 2330/5 (=466)
(c) A1 £21.6 (awrt)
(c) Answer: £21.6
(d) B1 Stall A cao
(d) B1 Stall A has the smaller standard deviation (£2.76 compared with £21.6), so even though both stalls have the same mean (£42), Stall A's daily sales vary less around it, oe
(d) Answer: Stall A, because it has the smaller standard deviation.

Question 8

(a) B1 1 cao
(a) Answer: 1
(b) M1 0, 7, 12, 9, 4, oe (correct fx products)
(b) A1 32 cso
(b) Answer: 32
(c) B1 1.6 cao (ft)
(c) Answer: 1.6
(d) M1 0, 7, 24, 27, 16, oe (correct fx² products)
(d) A1 74 cso
(d) Answer: 74
(e) M1 74/20 - 1.6² (=3.7-2.56), oe
(e) M1 = 1.14
(e) A1 1.07 (awrt)
(e) Answer: 1.07

Question 9

(a) B1 25 cao
(a) Answer: 25
(b) M1 20, 135, 275, 210, oe (correct fx products using midpoints)
(b) A1 640 cso
(b) Answer: 640
(c) B1 21.3 (cm) awrt (ft)
(c) Answer: 21.3 cm
(d) M1 100, 2025, 6875, 7350, oe (correct fx² products using midpoints)
(d) A1 16350 cso
(d) Answer: 16350
(e) M1 16350/30 (=545), oe
(e) M1 545 - 21.3333² (=545-455.111=89.9), ft their mean
(e) A1 9.48 (awrt)
(e) Answer: 9.48 cm

Question 10

(a) B1 the coded values are much smaller, single or two-digit numbers, which are easier to square and sum than the original 4-digit masses, oe
(a) Answer: The coded values are much smaller and simpler to work with than the original 4-digit masses.
(b) M1 4+11-1+8+2 (=24), oe
(b) A1 24/5 = 4.8
(b) Answer: 4.8
(c) B1 1024.8 (g), ft (1020+their 4.8)
(c) Answer: 1024.8 g
(d) M1 deviations -0.8, 6.2, -5.8, 3.2, -2.8, so sum of (y-mean)² = 90.8, oe
(d) M1 90.8/5 (=18.16)
(d) A1 4.26 (awrt)
(d) Answer: 4.26
(e) B1 4.26 (g), ft, because subtracting a constant from every value does not change how spread out the data is, oe
(e) Answer: 4.26 g

Question 11

(a) B1 one loaf from each oven might not be typical; a sample of several loaves reduces the effect of one unusually heavy or light loaf and gives a fairer comparison, oe
(a) Answer: A single loaf may not be typical; a sample of several loaves gives a fairer comparison.
(b) B1 weigh the same number of loaves from each oven, made using the same recipe and weighed on the same scales, oe
(b) Answer: Weigh equal-sized samples from each oven, made the same way and weighed on the same scales.
(c) B1 Oven A mean: (798+802+805+797+803)/5 = 4005/5 = 801
(c) B1 Oven B mean: (780+815+790+810+805)/5 = 4000/5 = 800
(c) M1 Oven A: deviations -3,1,4,-4,2, so sum of (x-mean)² = 46, oe
(c) M1 46/5 (=9.2)
(c) A1 Oven A: 3.03 g (awrt)
(c) M1 Oven B: deviations -20,15,-10,10,5, so sum of (x-mean)² = 850, oe
(c) M1 850/5 (=170)
(c) A1 Oven B: 13.0 g (awrt)
(c) Answer: Oven A: mean 801 g, SD 3.03 g. Oven B: mean 800 g, SD 13.0 g.
(d) B1 Oven A has the smaller standard deviation (3.03 g compared with 13.0 g), ft
(d) B1 so Oven A produces more consistent (less variable) loaf weights than Oven B, oe
(d) Answer: Oven A produces more consistent loaf weights, since its standard deviation (3.03 g) is much smaller than Oven B's (13.0 g).

Question 12

(a) B1 42 (m) cao
(a) Answer: 42 m
(b) M1 deviations 0,3,-2,-1, so sum of (x-mean)² = 14, oe
(b) M1 14/4 (=3.5)
(b) A1 √3.5 = 1.87 cso
(b) Answer: 1.87 m
(c) M1 recognises the new mean is still 42 m: (168+42)/5 = 42, oe
(c) M1 sum of (x-mean)² is still 14 (the new deviation is 0), so 14/5 (=2.8)
(c) A1 1.67 (awrt)
(c) Answer: 1.67 m
(d) B1 the new throw is equal to the mean, so it adds 0 to the sum of squared deviations, oe
(d) B1 but that sum is now shared between 5 throws instead of 4, so the average squared deviation (variance), and therefore the standard deviation, falls, oe
(d) Answer: The new throw equals the mean, adding nothing to the sum of squared deviations, but dividing by 5 instead of 4 reduces the variance and standard deviation.

Question 13

(a) B1 the student did not divide the sum of (x-mean)² by n (=5) before taking the square root; it should be √40/5, not √40, oe
(a) Answer: The student forgot to divide by n (5) before taking the square root.
(b) M1 40/5 (=8), oe
(b) A1 2.83 (awrt)
(b) Answer: 2.83

Question 14

(a) B1 45 (minutes) cao
(a) Answer: 45 minutes
(b) B1 212/8 = 26.5 cso
(b) Answer: 26.5 minutes
(c) M1 deviations -4.5,-1.5,-2.5,-3.5,18.5,-0.5,-2.5,-3.5, so sum of (x-mean)² = 402, oe
(c) M1 402/8 (=50.25)
(c) A1 √50.25 = 7.09 cso
(c) Answer: 7.09 minutes
(d) M1 26.5 - 2(7.09) and 26.5 + 2(7.09), oe
(d) A1 12.3 and 40.7 (awrt), both boundaries correct
(d) Answer: 12.3 minutes and 40.7 minutes
(e) B1 45 minutes, because 45 is greater than the upper boundary of 40.7 (ft), oe
(e) Answer: 45 minutes, since it is above the upper boundary of 40.7 minutes.