Topic 7 Statistics
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Transcript of Topic 7 Statistics
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TOPIC 7: STATISTICS *bring graph paper & french curve* *paper 2
Mean, Mode & Median of Ungrouped Data
mean,
f
xx mode = most frequent
median = middle
EXAMPLE 1 :Find the mode, median and mean for
3, 4, 0, 2, 2, 1, 5, 6, 7
Answer : Mode = 2 ascending orderMedian = 3 0, 1, 2, 2, 3, 4, 5, 6, 7
Mean ,
f
xx =
9
765432210
= 3.333
EXAMPLE 2 :
Find the mode, median and mean for2, 3, 2, 2, 6, 8, 9, 3, 2, 3
Answer : Mode = 2 ascending order
Median =2
33 2, 2, 2, 2, 3, 3, 3, 6, 8, 9
= 3
Mean ,
f
xx =
10
9863332222
= 4
1. Find the mode, median and mean for4, 5, 0, 2, 2, 1, 5, 5, 3
[Ans : Mode = 5, Median = 3, Mean = 3]2. Find the mode, median and mean for
52, 60, 70, 55, 56, 56
[Ans : Mode = 56, Median = 56, Mean = 58.167]3. If the mean of the set of data 4, k, 5, 6, 5, 6 and 4is 5, where k is a constant, find the mode.
[Ans : Mode = 5]Mean, Mode & Median of Grouped Data
mean,
f
xfx
mode = using histogram
median =c
f
Ff
Lm
2
1
L = lower boundary of the median class
f= total frequency (jumlah kekerapan)F= cumulative frequency before the median class
c = size of class interval(selang kelas)fm = frequency of the median classRange(julat) = highest value - lowest value
First quartile, Q1 = thf4
1 observation
=
1
1
1
41
Q
mQ
Q cf
Ff
L
Third quartile, Q3 = thf4
3 observation
=
3
3
3
4
3
Q
mQ
Q cf
Ff
L
Interquartile range = Q3Q1(julat antara kuartil)
EXAMPLE 3 :
Find the mode, median and mean for the data inthe table below.
Answer : Mode = 3 Median = thf2
1 observation
= 20th observation
= 3
Mean ,
f
xfx =
40
107
= 2.675
Score 0 1 2 3 4 5
Frequency 2 5 8 17 5 3
Score,x 0 1 2 3 4 5
Frequency,f 2 5 8 17 5 3 f 40fx 0 5 16 51 20 15 fx 107
(min) (mod)
(sempadan
bawah)
highest frequency
for class interval
Modal class =
(kuartil pertama)
(kuartil ketiga)
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4. Find the mode, median and mean for the data in
the table below.
[Ans : Mode = 4, Median = 6, Mean = 7.2]
Score 2 4 6 9 12 13
Frequency 1 3 2 1 2 1
5. The table below shows the scores obtained
by number of children in a game. If the meanscore is 27, calculate the value of p.
[Ans : p = 3]
Score Frequency
20 5
30 p
40 2
Mode from Histogram
EXAMPLE 4 :Table 1 shows the frequency distribution of thescores of a group of pupils in a game.
Score Number of pupils
1 - 10 5
11 - 20 8
21 - 30 12
31 - 40 441 - 50 6
51 - 60 5
Table 1(a) Use the graph paper to answer this question.Using a scale of 2 cm to 10 scores on thehorizontal axis and 2 cm to 2 pupils on the vertical
axis, draw a histogram to represent the frequencydistribution of the scores.(b) Find the mode score.
Answer : (a)
Score Number of pupils Midpoint1 - 10 5 5.5
11 - 20 8 15.5
21 - 30 12 25.5
31 - 40 4 35.5
41 - 50 6 45.5
51 - 60 5 55.5
(b) Mode score = 24
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6. Table 2 shows the frequency distribution of the
scores of a group of pupils in a test.
Score Number of pupils
50 - 54 2
55 - 59 3
60 - 64 7
65 - 69 9
70 - 74 4
75 - 79 1Table 2
(a) Use the graph paper to answer this question.Using a scale of 2 cm to 5 scores on the horizontalaxis and 2 cm to 1 pupils on the vertical axis, draw
a histogram to represent the frequency distributionof the scores.
(b) Find the mode score.
[Ans : Mode = 66]
7. Table 3 shows the frequency distribution of the
scores of a group of pupils in a game.
Score Number of pupils
65 - 69 2
70 - 74 8
75 - 79 16
80 - 84 12
85 - 89 6
90 - 94 495 - 99 2
Table 3(a) Use the graph paper to answer this question.
Using a scale of 2 cm to 5 scores on the horizontalaxis and 2 cm to 2 pupils on the vertical axis, draw ahistogram to represent the frequency distribution ofthe scores.(b) Find the mode score.
[Ans : Mode = 78]EXAMPLE 5 :Find the modal class and calculate the median for the data in the table below.
Marks 40 - 45 46 - 51 52 - 57 58 - 63 64 - 69 70 - 75 7681
Frequency 3 6 8 14 10 5 4
Answer :
Marks Frequency Upper Boundary Cumulative Frequency
40 - 45 3 45.5 3
46 - 51 6 51.5 9
52 - 57 8 57.5 17
58 - 63 14 63.5 31
64 - 69 10 69.5 41
70 - 75 5 75.5 46
76 - 81 4 81.5 50
Modal class = 5863 Median =c
f
Ff
Lm
2
1
= thf2
1 observation
= 25th observation
=6
14
17)50(2
1
5.57
= 57.5 + 3.429= 60.929
L = 57.5
f= 50F= 17
c = 63.5 - 57.5 = 6
m = 14
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8. Find the modal class and calculate the median 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
[Ans : Modal class = 46-60, Median = 48.147]9. Find the modal class and calculate the medianfor the data in the table below.
[Ans : Modal class = 10-14, Median = 10.75]
Points 0 - 4 5 - 9 10 - 14 15 - 19
frequency 2 5 8 3
10. The table below shows the marks obtained by34 students in a SPM trial examination forMathematics. Find the modal class and calculate the
median mark.
[Ans : Modal class = 41-60, Median mark = 55.167]
Marks Number of students
01 - 20 2
21 - 40 4
41 - 60 1561 - 80 10
81 - 100 3
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Ogive
EXAMPLE 6 :Table below shows the frequency distribution of the ages of a group of workers in an office.
Age(years) 20 - 24 25 - 29 30 - 34 35 - 39 40 - 44 45 - 49 5054
Number of workers 2 3 7 14 9 4 1
(a) Use the graph paper to answer this question.Using a scale of 2 cm to 5 years on the horizontal axis and 2 cm to 5 workers on the vertical axis, draw an
ogive to represent the frequency distribution of the ages.
(b) From the ogive, find the(i) median ages of the workers.
(ii) interquartile range.
Answer :
Age(years) Number of workers Upper Boundary Cumulative Frequency
< 20 0 19.5 0
20 - 24 2 24.5 2
25 - 29 3 29.5 5
30 - 34 7 34.5 12
35 - 39 14 39.5 26
40 - 44 9 44.5 35
45 - 49 4 49.5 39
5054 1 54.5 40
(a)
(b) (i) Median = thf2
1 observation (ii) Q3 = thf4
3 observation Q1 = thf4
1 observation
= 20th observation = 30th observation = 10th observation= 37.75 = 41.25 = 33.75
Interquartile range = Q3Q1= 41.25 - 33.75= 7.5
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11. Table below shows the frequency distribution of the score of a group of students in a game.
Score 5 - 9 10 - 14 15 - 19 20 - 24 25 - 29 30 - 34 3539
Number of students 2 3 7 12 14 9 3
(a) Use the graph paper to answer this question.Using a scale of 2 cm to 5 score on the horizontal axis and 2 cm to 5 students on the vertical axis, draw an
ogive to represent the frequency distribution of the scores.(b) From the ogive, find the
(i) median scores of the students.(ii) interquartile range.
[Ans : (i) Median scores = 25 (ii) Interquartile range = 9.5]
12. Table below shows the frequency distribution of the score of a group of students in a game.
Score 1 - 10 11 - 20 2130 31 - 40 41 - 50 51 - 60 6170 7180 8190 91100
Number ofstudents
3 5 7 12 15 19 16 10 8 5
(a) Use the graph paper to answer this question.Using a scale of 2 cm to 20 scores on the horizontal axis and 2 cm to 10 students on the vertical axis,draw an ogive to represent the frequency distribution of the scores.(b) From the ogive, find the
(i) median scores of the students.
(ii) interquartile range.
[Ans : (i) Median scores = 54.5 (ii) Interquartile range =2 9]
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Measures of Dispersion for Ungrouped Data
Variance,
22
2
f
x
f
x
Standard deviation, variance
EXAMPLE 7 :22, 20, 26, 25, 28, 30, 38, 35, 45, 40, 32
From the data given above, find the(a) range (d) mean
(b) median (e) variance(c) interquartile range (f) standard deviation
Answer : ascending order20, 22, 25, 26, 28, 30, 32, 35, 38, 40, 45
Q1 Q2 Q3(a) range = 45 20 = 25(b) median = 30(c) interquartile range = Q3Q1 = 3825 = 13
(d) mean,
f
xfx
=11
4540383532302826252220
= 31
(e) variance,
22
2
f
x
f
x
=11
454038353230282625222022222222222
-
2
114540383532302826252220
= 1017 - 312= 56
(f) standard deviation, variance
= 56
= 7.483
13. 5, 1, 2, 3, 4, 6, 3, 8, 2, 5, 9
From the data given above, find the(a) range (d) mean(b) median (e) variance
(c) interquartile range (f) standard deviation
[Ans : (a)Range = 8 (b)Median = 4 (c)Interquartile range =4
(d)Mean = 4.364 (e)Variance = 5.868 (f)Standard deviation = 2.422]
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EXAMPLE 8 :
12, 8, 10, 6, 6, 7, 13, 5, 8, 10, 13, 15From the data given above, find the(a) range (d) mean
(b) median (e) variance(c) interquartile range (f) standard deviation
Answer : ascending order
5, 6, 6, 7, 8, 8, 10, 10, 12, 13, 13, 15
Q1 Q2 Q3(a) range = 15 5 = 10
(b) median =2
108 = 9
(c) interquartile range = Q3Q1
=2
1312 2
76
= 12.56.5
= 6
(d) mean,
f
xf
x
=12
151313121010887665
= 9.417
(e) variance,
22
2
f
x
f
x
=12
151313121010887665222222222222
-
2
12
151313121010887665
= 117.1679.4172
= 28.487
(f) standard deviation, variance
= 487.28 = 5.337
14. 12, 17, 13, 19, 15, 8, 12, 11
From the data given above, find the(a) range (d) mean(b) median (e) variance
(c) interquartile range (f) standard deviation
[Ans : (a)Range = 11 (b)Median = 12.5 (c)Interquartile range =4.5
(d)Mean = 13.375 (e)Variance = 10.734 (f)Standard deviation = 3.276]
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Measures of Dispersion for Grouped Data
Variance,
22
2
f
fx
f
fx
Standard deviation, variance
EXAMPLE 9 :Calculate the interquartile range, mean, variance and the standard deviation for the data.
Class interval 6 - 8 9 - 11 12 - 14 15 - 17 18 - 20
Frequency 3 8 14 11 4
Answer :
Classinterval
Frequency UpperBoundary
CumulativeFrequency
< 6 0 5.5 06 - 8 3 8.5 39 - 11 8 11.5 11
12 - 14 14 14.5 25
15 - 17 11 17.5 3618 - 20 4 20.5 40
Q3 = thf4
3 observation Q1 = thf4
1 observation
= 30th observation = 10th observation
=3
3
3
4
3
Q
mQ
Q cf
Ff
L
=
1
1
1
4
1
Q
mQ
Q cf
Ff
L
=3
11
25)40(4
3
5.14
=
38
3)40(4
1
5.8
= 14.5 + 1.364 = 8.5 + 2.625
= 15.864 = 11.125
Interquartile range = Q3Q1= 15.86411.125= 4.739
Classinterval
Frequency,f
Midpoint,x
fx fx2
6 - 8 3 7 21 1479 - 11 8 10 80 800
12 - 14 14 13 182 236615 - 17 11 16 176 281618 - 20 4 19 76 1444
f= 40 xf = 2xf =535 7573
L = 14.5
f= 40F= 25c = 17.5 - 14.5 = 3
m = 11
L = 8.5
f= 40F= 3c = 3
m = 8
Mean,
f
xfx =
40
535= 13.375
Variance,22
2
f
fx
f
fx
=40
7573 -
2
40
535
= 189.325178.891
= 10.434
Standard deviation, variance = 434.10 = 3.23
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15. Calculate the interquartile range, mean, variance and the standard deviation for the data.
Class interval 1 - 5 6 - 10 11 - 15 16 - 20 21 - 25
Frequency 3 13 23 9 2
[Ans : Mean = 12.4, 2 = 20.64, = 4.543]
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16. Calculate the interquartile range, mean, variance and the standard deviation for the data.
Class interval 1 - 3 4 - 6 7 - 9 10 - 12 13 - 15
Frequency 1 3 8 6 2
[Ans : Mean = 8.75, 2 = 8.888, =2.981]
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17. Calculate the interquartile range, mean, variance and the standard deviation for the data.
Class interval 10 - 19 20 - 29 30 - 39 40 - 49
Frequency 5 7 5 3
[Ans : Mean = 27.5, 2 = 101, = 10.05]