7th Maths : Term 3 Unit 5 : Statistics : Text Book Back Exercises Questions with Answers, Solution

**Exercise
5.1**

**1. Fill in the blanks.**

(i) The mean
of first ten natural numbers is _______.**[Answer:
5.5]**

(ii) If the
average selling price of 15 books is ₹ 235, then the total selling price is
_______.**[Answer:
3,525]**

(iii) The
average of the marks 2, 9, 5, 4, 4, 8, 10 is ________.** [Answer: 6]**

(iv) The
average of integers between –10 to 10 is _______.** [Answer: 0]**

**2. Ages of 15 students in 8th standard
is 13, 12, 13, 14, 12, 13, 13, 14, 12, 13, 13, 14, 13, 12, 14. Find the mean age
of the students.**

**Solution: **

Arithmetic Mean = Sum of all observations / Number of observations

= [13 + 12 + 13 + 14 + 12 + 13 + 13 + 14 + 12 + 13 + 13 + 14 +
13 + 12 + 14] / 15

= 195 / 15 = 13

Mean age of the students = 13

** **

**3. The marks of 14 students in a science
test out of 50 are given below. 34, 23, 10, 45, 44, 47, 35, 37, 41, 30, 28, 32,
45, 39 Find**

**(i) the mean mark.**

**(ii) the maximum mark obtained.**

**(iii)
the minimum mark obtained.**

**Solution: **

(i) Mean marks = Sum of all marks / Total number of marks

= [ 34 + 23 + 10 + 45 + 44 + 47 + 35 + 37 + 41 + 30 + 28 + 32 +
45 + 39 ] / 14

= 490 / 14

Mean marks = 35

**(ii)** Maximum mark obtained = 47

**(iii)** Minimum mark obtained = 10

** **

**4. The mean height of 11 students in
a group is 150 cm. The heights of the
students are 154 cm, 145 cm, Y cm, Y + 4 cm, 160 cm,
151 cm, 149 cm, 149 cm, 150 cm, 144 cm and 140
cm. Find the value of Y and the heights of two students?**

**Solution:**

Mean Height = Sum of heights of all students / Number of
students

150 = [ 154+ 145 + Y + ( Y+ 4) + 160 + 151 + 149 + 149 + 150 +
144 + 140 ] / 11

150 = [ 1342 + Y + Y + 4 ] / 11

150 = [ 1346 + 2Y ] / 11

150 × 11 = 1346 + 2Y

1650 = 1346 + 2Y

2Y = 1650 – 1346 = 304

Y = 304 / 2 = 152

Height of two students are Y and Y + 4

=> 152 and 152+4

=> 152 cm and 156 cm

** **

**5. The mean of runs scored by a cricket
team in the last 10 innings is 276. If the scores are 235, 400, 351, x, 100,
315, 410, 165, 260, 284, then find the runs scored in the fourth innings.**

**Solution:**

Let the runs scored in the fourth innings be *x*.

Mean runs scored = Total runs of all innings / number of innings

276 = [ 235 + 400 + 351 + *x*
+ 100 + 315 + 410 + 165 + 260 + 284 ] / 10

276 = [ 2520 + *x* ] /
10

276 × 10 = 2520 + *x*

2760 = 2520 + *x*

*x* = 2760 – 2520 = 240

∴ Number of runs scored in the fourth innings = 240

** **

**6. Find the mean of the following data.
5.1, 4.8, 4.3, 4.5, 5.1, 4.7, 4.5, 5.2, 5.4, 5.8, 4.3, 5.6, 5.2, 5.5**

**Solution:**

Mean = Sum of all numbers / Number of values

= [ 5.1+ 4.8 + 4.3 + 4.5 + 5.1 + 4.7 + 4.5 + 5.2 + 5.4 + 5.8 +
4.3 + 5.6 + 5.2 + 5.5 ] / 14

= 70.0 / 14 = 5

Mean = 5

** **

**7.
Arithmetic mean of 10 observations was found to be 22. If one more observation 44
was to be added to the data, what would be the new mean?**

**Solution:**

Arithmetic mean of 10 observation is 22.

Arithmetic mean = Sum of all observations / Number of
observations

22 = sum of 10 observations / 10

Sum of 10 observations = 22 × 10 = 220

Now if new number is added, then

Mean of 11 observations = [ Sum of 10 observation + 44 ] / 11

= [220 + 44] / 11= 264 / 10 = 24

New mean = 24

** **

__Objective type questions__

**8.
________ is a representative value of the entire data. **

(i) Mean

(ii) range

(iii) minimum value

(iv) maximum value

**[Answer: (1) Mean]**

**9.
The mean of first fifteen even numbers is _______ **

(i) 4

(ii) 16

(iii) 5

(iv) 10

**[Answer: (2) 16]**

**Solution
: **[2 + 4 + 6 +.... + 30] / 15
= 2[ 1+ 2 + 3 + ....15] / 15 = 2 × [120 / 15] = 16

**10.
The average of two numbers are 20. One number is 24, another number is ________**

(i) 16

(ii) 26

(iii) 20

(iv) 40

**[Answer: (1) 16]**

**Solution :** [*x + y*]* / *2 = 20

*x + y* = 40

24+* y* = 40

*y* = 40 – 24 = 16

**11.
The mean of the data 12, ***x***, 28
is 18. Find the value of ***x***.**

(i) 18

(ii) 16

(iii) 14

(iv) 22

**[Answer: (3) 14]**

**Solution: [**12 + *x *+ 28] / 3 = 18

*x* + 40 = 54

*x* = 14

__ANSWERS:__

**Exercise 5.1**

1. (i) 5.5 (ii) 3,525
(iii) 6 (iv) 0

2. 13

3. (i) 35 (ii) 47
(iii) 10

4. Y=152 ; Height of
two students are 152* cm *and 156 *cm*

5. 240

6. 5

7. 24

**Objective type questions:**

8. (i) Mean

9. (ii) 16

10. (i) 16

11. (iii) 14

Tags : Statistics | Term 3 Chapter 5 | 7th Maths , 7th Maths : Term 3 Unit 5 : Statistics

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7th Maths : Term 3 Unit 5 : Statistics : Exercise 5.1 | Statistics | Term 3 Chapter 5 | 7th Maths

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