7 STATISTICS.doc

8. STATISTICS

1. Measures of Central Tendency

Mode, median and meanFor a sample of discrete data, the mode is the observation, x with the highest frequency, f.

For grouped data in a cumulative frequency table, the median =

where L is the lower boundary of the median classN is total frequencyF is the cumulative frequency before the median classC is the interval of the median classfm is the frequency of the median class

The mean is the average of all the observations. Hence, . For grouped data, .

(a) Find the mode, median and mean for 2, 3, 1, 2, 6, 8, 9, 3, 2, 3.

[Mode = 2, Median = 3, Mean = 4]

(b) Find the mode, median and mean for the data in the table below.

[Mode = 4, Median = 6, Mean = 7.2](c) Find the modal class and calculate the median

and mean for the data in the table below.

[Modal class = 10-14, Median = 10.75, Mean = 10.33]

(d) Find the modal class and calculate the median and mean for the data in the table below.

Marks 1 - 15 16 - 30 31 - 45 46 - 60 61 - 75 76 - 90

f 8 11 25 34 16 6

Median =

= 46.55

[Modal class = 46-60, Median = 48.15, Mean = 46.53]

(e) Estimate the mode and calculate the median and (f) Estimate the mode and calculate the median and

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Score 2 4 6 9 12 13Frequency 1 3 2 1 2 1

Points 0 - 4 5 - 9 10 - 14 15 - 19frequency 2 5 8 3

mean for the histogram below.

[Mode = 72.8, Median = 72.71, Mean = 72.6]

mean for the histogram below.

[Mode = 152.5, Median = 150.09, Mean = 149.12](g) Find the modal class and find the median and (h) Find the modal class and find the median and

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mean for the ogive below.

Class 5-9

10-14

15-19

20-24

25-29

30-34

35-39

4 18 28 16 8 4 2

Modal class = 15 - 19

[Modal class = 15-19, Median = 17.6, Mean = 18.625]

mean for the ogive below.

[Modal class = 45-49, Median = 49.5, Mean = 50.25]

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2. Measures of Dispersion:

Range, quartiles, interquartile range, variance and standard deviation.

The range of a sample is the difference between the observations with the highest value and the lowest value.

The first quartile, Q1 = and the third quartile, Q3 =

Interquartile range = Q3 – Q1. It should be recognised that the median is Q2.

The variance, for discrete data.

For grouped data, .

The standard deviation is actually

For discrete data, .

For grouped data, .

(a) Find the range and the interquartile for 5, 1, 2, 3, 4, 6, 3, 8, 2, 5, 9.

[Range = 8, Interquartile range = 4]

(b) Find the range and the interquartile range for12, 17, 13, 19, 15, 8, 12, 11.

[Range = 11, Interquartile range = 4.5](c) Find the range and the interquartile for

Score 1 4 5 6 8 9Frequency 1 3 1 1 2 1

[Range = 8, Interquartile range = 4]

(d) Find the range and the interquartile range forPoints 2 4 6 8

No. of person 3 5 2 2

[Range = 6, Interquartile range = 3]

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(e) Find the interquartile range for the table below.Marks 1 - 20 21 - 40 41 - 60 61- 80 81 - 100

No. of Students 4 9 12 10 5

[Interquartile range = 40.77 marks]

(f) Find the interquartile range for the table below.Age (year) 1-20 21-40 41-60 61-80

No. of residents 66 99 57 28

[Interquartile range = 28.95 years](e) Calculate the interquartile range for the ogive

below.

[Interquartile range = 12.5 mm]

(f) Calculate the interquartile range for the ogive below.

[Interquartile range = 6.5 years]

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(g) Calculate the interquartile range for the histogram below.

[Interquartile range = 5.295 kg]

(h) Calculate the interquartile range for the histogram below.

[Interquartile range = 18.03 minutes]

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(a) Find the mean, variance and the standard deviation for the data below.5, 12, 6, 3, 6, 10.

Mean= = 7

x 5 12 6 3 6 10x-mean -2 5 -1 -4 -1 3(x – mean)2

4 25 1 16 1 9

= 4 +25

[Mean = 7, 2 = 9.333, = 3.055]

(b) Find the mean, variance and the standard deviation for the data below.18, 12, 16, 11, 19, 18, 12, 14.

[Mean = 15, 2 = 7.5, = 2.739](c) Complete the table below and calculate the

mean, variance and the standard deviation for the data.

x 1 2 3 4 5 6

f 1 3 4 7 3 2 20

fx 1 6 12 28 15 12 74

2.7 1.7 0.7 -0.3 -1.3 -2.3

7.29 2.89 0.49 0.09 1.69 5.29

7.29 8.67 1.96 0.63 5.07 10.58 34.2

Mean =

Variance = = 1.71

Standard deviation = = 1.308

[Mean = 7, 2 = 9.333, = 3.055]

(d) Complete the table below and calculate the mean, variance and the standard deviation

for the data.x 2 4 6 8 10 12

f 1 2 2 2 1 2 10

fx 2 8 12 16 10 24 72

5.2 3.2 1.2 -0.8 -2.8 -4.8

27.04

10.24

1.44 0.64 7.8423.0

427.0

420.4

82.88 1.28 7.84

46.08

105.6

[Mean = 7.2, 2 = 10.56, = 3.25]

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(e) Complete the table below and calculate the mean, variance and the standard deviation

for the data.Class 1 - 3 4 - 6 7 - 9 10 - 12 13 - 15

f 1 3 8 6 2 20

x 2 5 8 11 14

fx 2 15 64 66 28 175

x2 4 25 64 121 196

4 75 512 726 392 1709

Variance =

- (8.75)2 = 8.888

[Mean = 8.75, 2 = 8.8875, = 2.981]

(f) Complete the table below and calculate the mean, variance and the standard deviation

for the data.Class 10 - 19 20 - 29 30 - 39 40 - 49

f 5 7 5 3

x

fx

x2

[Mean = 27.5, 2 = 101, = 10.05]

(g) Complete the table below and calculate the mean, variance and the standard deviation

for the data.Class 1 - 5 6 - 10 11 - 15 16 - 20 21 - 25

f 3 13 23 9 2

x

fx

x2

[Mean = 12.4, 2 = 20.64, = 4.543]

(h) Complete the table below and calculate the mean, variance and the standard deviation

for the data.Class 0 - 19 20 - 39 40 - 59 60 - 79

f 4 7 6 3

x

fx

x2

[Mean = 37.5, 2 = 376, = 19.39]

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(i)

Extract the data from the histogram above and complete the table below. Calculate the mean, variance and the standard deviation for the data.

Class x f fx

[Mean = 55.55, 2 = 41.6475, = 6.453]

(j)

Extract the data from the histogram above and complete the table below. Calculate the mean, variance and the standard deviation for the data.

Class x f fx

[Mean = 150.3, 2 = 172.36, = 13.13]

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1. SPM 2003 Paper 2 Compulsory Question No. 5A set of test marks: x1, x2, x3, x4, x5, x6 has a mean value of 5 and standard deviation of 1.5.(a) Find the

(i) sum of marks, x,

(ii) sum of squared marks, x2. [ 3 ](b) Each mark is multiplied by 2 and then 3 is added.

For the set of new marks, find the(i) mean,

(ii) variance. [ 4 ]

Answer: (a) (i) 30 (ii) 163.50 (b) (i) 13 (ii) 3_________________________________________________________________________________________

2. SPM 2004 Paper 2 Compulsory Question No. 4A set of data contains 10 numbers. The sum of these numbers is 150and the sum of their squares is 2472.(a) Find the mean and variance for these 10 numbers. [ 3 ](b) Another number is added to this set of data and the mean increase by 1.

Find(i) the value of this number,

(ii) the standard deviation for the set of 11 numbers. [ 4 ]

Answer: (a) mean = 15, variance = 22.2 (b) (i) 26 (ii) 5.494_________________________________________________________________________________________

3. SPM 2005 Paper 1 Compulsory Question No. 23

The mean of four numbers is .

The sum of the squares of the numbers is 100 and the standard deviation is 3k.Express m in terms of k. [ 3 ]

Answer: 25 - 9k2

_________________________________________________________________________________________

4. SPM 2005 Paper 2 Compulsory Question No. 4

The diagram beside is a histogram which represents thedistribution of the marks obtained by 40 pupils in a test.(a) Without using an ogive,

calculate the median mark. [ 3 ](b) Calculate the standard deviation of the distribution. [ 4 ]

Answer: (a) 24.07 (b) 11.74_________________________________________________________________________________________

5. SPM 2006 Paper 1 Compulsory Question No. 24A set of positive integers consists of 2, 5 and m.The variance for this set of integers is 14.Find the value of m.

[Answer: 11][ 4 ]

[4 marks]

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Number of Pupils

Marks0

2

4

68

10

12

14

0.5 10.5 20.5 30.5 40.5 50.5

6. SPM 2006 Paper 2 Compulsory Question No. 6The table beside shows the frequency distribution of the scores of a groupof pupils in a game.(a) It is given that the median score of the distribution is 42.

Calculate the value of k. [ 3 ](b) Use the graph paper provided by the invigilator to answer this question.

Using a scale of 2 cm to 10 scores on the horizontal axis and 2 cm to2 pupils on the vertical axis, draw a histogram to represent thefrequency distribution of the scores.Find the mode score. [ 4 ]

(c) What is the mode score if the score of each pupil is increased by 5? [ 1 ]

Answer: (a) 4 (b) 42.6 (c) 47.6_________________________________________________________________________________________

7. SPM 2007 Paper 1 Compulsory Question No. 22A set of data consists of five numbers. The sum of the numbers is 60 andthe sum of the squares of the numbers is 800. Find, for the five numbers(a) the mean,(b) the standard deviation. [ 3 ]

Answer: (a) 12 (b) 4_________________________________________________________________________________________

8. SPM 2007 Paper 2 Compulsory Question No. 5The table below shows the cumulative frequency distribution for the scores of 32 students in a competition.

(a) Based on the table above, copy and complete the table below. [ 1 ]Score 0 - 9 10 - 19 20 - 29 30 - 39 40 - 49

Number of students

(b) Without drawing an ogive, find the interquartile range of the distribution. [ 5 ]

Answer: (a) 4, 6, 10, 8, 4 (b)

_________________________________________________________________________________________

9. SPM 2008 Paper 1 Compulsory Question No. 22A set of seven numbers has a mean of 9.(a) Find .(b) When a number k is added to this set, the new mean is 8.5. Find the value of k. [ 3 ]

Answer: (a) 63 (b) 5_________________________________________________________________________________________

10. SPM 2008 Paper 2 Compulsory Question No. 6The table beside shows the marksobtained by 40 candidates in a test.Given that the median mark is 35.5,find the value of x and of y.Hence, state the modal class. [ 6 ]Answer: x = 13, y = 5, modal class = 20 - 29

________________________________________________________________________________________

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ScoreNumber of Pupils

10 - 1920 - 2930 - 3940 - 4950 - 5960 - 69

12812k1

Score < 10 < 20 < 30 < 40 < 50Number of students 4 10 20 28 32

Marks Number of Candidates10 - 1920 - 2930 - 3940 - 4950 - 59

1xy

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