Samacheer Class 7 Maths - Statistics Intext Questions

Height of 15 students in our class.
130cm, 150 cm, 155 cm, 142 cm, 138 cm, 145 cm, 148 cm, 147 cm, 148cm, 143 cm, 141cm, 152 cm, 147 cm, 139 cm, 155 cm.
Ascending order:
130cm, 138cm, 139cm, 141cm, 142cm, 143cm, 145cm, 147cm, 147cm, 148cm, 148 cm, 150cm, 152 cm, 155cm, 155cm.

Try These (Text book Page No. 97)
Find the Arithmetic Mean or average of the following data.

Given number are 3, 6, 9, 12, 15

If mean is increased by 2 then,

Sum of observations = 5 × 11 = 55
Difference in sum = 55 – 45 = 10
∴ Each number is increased by 2 if the mean is increased by 2.

If the first two items is increased by 3, then the numbers will be 3 + 3, 6 + 3 ⇒ 6, 9.
If last two numbers are decreased by 3, then the numbers will be 12 – 3, 15 – 3 ⇒ 9,12.

= \(\frac { 45 }{ 5 } \) = 9
There is no change in the mean.

Exercise 5.2
Try These (Text book Page No. 101)

Arranging the numbers in ascending order we get 0, 2, 2, 2, 3, 3, 3, 4, 4, 5, 5, 6
Since 2 and 3 occurs the maximum of 3 times. So mode of this data is 2 and 3.

Arranging the given data in ascending order : 3, 3, 3, 4, 9,11, 12, 12,12, 15, 19.
The data 3 and 12 occurs the maximum of 3 times.
So mode of this data is 3 and 12.

Odd number between 20 to 40 are 21, 23, 25, 27, 29, 31, 33, 35, 37, 39.
As all numbers occurs only once there is no mode for this data.
Think (Text book Page No. 103)

Arranging in ascending order: 6, 7, 7, 9, 10, 11, 11, 12, 14, 14
Here n = 10, which is even.


Think (Text book Page No. 108)
Complete the table given below and observe it to answer the following questions.

If 99 becomes 0 or 200 becomes 101 then mean becomes 100.
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150 = \(\frac { 1342+Y+Y+4 }{ 11 } \)
150 = \(\frac { 1346+2Y }{ 11 } \)
150 × 11 = 1346 + 2Y
1650 = 1346 + 2Y
2Y = 1650 – 1346 = 304
Y = \(\frac { 304 }{ 2 } \) = 152
Height of two students are Y and Y + 4
⇒ 152 and 152 + 4
⇒ 152 cm and 156 cm

Let the runs scored in the fourth innings be x.

276 × 10 = 2520 + x
2760 = 2520 + x
x = 2760 – 2520 = 240
∴ Number of runs scored in the fourth innings = 240

Arithmetic mean of 10 observation is 22.

= \(\frac { 220+44 }{ 11 } \) = \(\frac { 264 }{ 11 } \) = 24
New mean = 24

Objective Type Questions

(iii) 14
Hint:
\(\frac { 12+x+28 }{ 3 } \) = 18
x + 40 = 54
x = 14
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Arranging the given data in ascending order 2, 2, 2, 2, 2, 3, 4, 4, 4, 5, 5, 6, 6, 7, 7, 8, 10.
Here the number 2 occurs 5 times which is the maximum
∴ Mode of this data is 2.

Arranging the given data in ascending order
27, 28, 28, 29, 29, 30, 31, 31, 31, 31, 31, 32, 33, 34, 34, 35, 35, 36, 38, 38.
Here the number 31 occurs 5 times which is the maximum.
∴ Mode of this data is 31.

Arranging the ages is ascending order: 25, 25, 25, 26, 36, 36, 36, 36, 38, 38, 39, 40.
The ages (in years) of 11 cricket players are given below. 25, 36, 39,38 40, 36, 25, 25, 38, o 26,36. Find the mode of the ages.
Arranging the ages is ascending order: 25, 25, 25, 26, 36, 36, 36, 36, 38, 38, 39, 40.
25 and 36 occurs maximum number of times.
∴ Mode is 25 and 36.

Arranging the given data in ascending order 18, 25, 28, 34, 35, 36, 45.
Here the number of observations n = 7, which is odd.

Hence Median = 34

Arranging the given data in ascending order 3, 4, 4, 5, 6, 6, 7, 7, 8, 10, 10, 12, 13, 16.
Here number of data n = 14, which is even


∴ Median = 7

Arranging the given 10 observations in ascending order 23, 25, 28, 33, 36, 39, 45, 45, 54, 56.
Here number of data n = 10, which is even

∴ Median = 37.5
If 35 is added to the above data then it will be the 5th term then number of data n = 11, odd
∴ Median = (\(\frac { n+1 }{ 2 } \)) th term = (\(\frac { 11+1 }{ 2 } \)) th term
= (\(\frac { 12 }{ 2 } \)) th term = 6th term
New median = 36

Objective Type Questions

(i) 6
Hint:

Median = \(\frac { 5+7 }{ 2 } \)
= \(\frac { 12 }{ 2 } \)
= 6
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Arranging the data in ascending order: 1, 1, 2, 2, 2, 3, 3, 3, 5, 5, 5
Here 2, 3 and 5 occurs 3 times each.
Which is the maximum number of times.
∴ Mode is 2, 3 and 5.

Arranging the given data in ascending order: 1, 1, 2, 2, 2, 3, 3, 4, 4, 6, 6
Clearly 2 occurs at the maximum of 3 times and so mode = 2
Here number of data of data n = 11, odd.
∴ Median = (\(\frac { n+1 }{ 2 } \)) th term
= (\(\frac { 11+1 }{ 2 } \)) th
term = (\(\frac { 12 }{ 2 } \)) th term
= 6 th term
Median = 3

Arranging is ascending order : 2, 4, 6, 11, 12, 13

Mean = \(\frac { 48 }{ 6 } \) = 8
All observation occurs only once and so there is no mode for this date.

Challenge Problems

Sum of observation = 6 × 8 = 48
If one more mark is added then number of observations = 6 + 1 = 7
Let the number be x
Still average = 8
∴ 8 = \(\frac { 48+x }{ 7 } \)
48 + x = 7 × 8
48 + x = 56
48 + x = 56 – 48
x = 8
∴ The number that is added = 8

Arranging in ascending order: 10, 10, 15, 21, 22, 24

Here n = 6, even

∴ Median = 18
Clearly the data 10 occurs maximum number of times and so 10 is the mode.
∴ Mode = 10

Arranging the data is ascending order: -8, -3, -2, -1, 0, 7, 13, 14
Here number of data n = 8, even
∴ Median

∴ Median = – 0.5

First 10 prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29

Mean = 12.9
Mean of first 10 prime numbers = 12.9
First 10 composite numbers are 4, 6, 8, 9, 10, 12, 14, 15, 16, 18

Mean of first 10 composite numbers = 11.2
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