Exercise 17.1

General Section

1 a) If $\sum fx$ = 450 and N = 15, find $\overline{X}$.

Solution:
We know,
$\overline{X} = \dfrac{ \sumfx}{N}$

$= \dfrac{450}{15}$
$= 30$

2 a) If $\overline{X}$ = 35 and $\sum fx$ = 455, find the number of terms (N).

Solution,
We know,
$N = \dfrac{ \sum fx}{\overline{X}}$

$= \dfrac{455}{35}$
$= 13$

3 a) If $\overline{X}$= 25, N = 10 $\sum fx$ = p, find the value of p.

Solution:
We know,

$\sum fx = \overline{X} × N$

$or, p = 25 × 10$
$\therefore p = 250$

4 a) In a series, if $\sum fm$=150 + 5a, $\overline{X}$=10 and $\sum f$= 10 + a, find the value of a.

Solution:
We know,
$\overline{X} = \dfrac{ \sum fx}{ \sum f}$

$or, 10 = \dfrac{150 + 5a}{10 + a}$

$or, 10(10+a) = 150 + 5a$

$or, 100 + 10a = 150 + 5a$
$or, 10a - 5a = 150 - 100$

$or, 5a = 50$
$or, a = \dfrac{50}{5}$
$\therefore a = 10$

4 c) If mean (\overline{X}) = 12, $\sum$fm = 70 + 10a and the number of frequency (N) = 5 + a, find the value of a.

Solution:
We know,
$\overline{X} = \dfrac{ \sum fm}{N}$

$or, 12 = \dfrac{70 + 10a}{5+a}$

$or, 12(5+a) = 70 + 10a$
$or, 60 + 12a = 70 + 10a$
$or, 12a - 10a = 70 - 60$
$or, 2a = 10$
$\therefore a = 5$

4 e) If $\overline{X}$ = 15, $\sum$fm = 350 + 13p and number of terms (N) = 32, find the value of p.

Solution:
We know,
$\overline{X} = \dfrac{ \sum fm}{N}$

$or, 15 = \dfrac{350 + 13p}{32}$

$or, 14 × 32 = 350 + 13p$
$or, 350 + 13p = 480$
$or, 13p = 480 - 350$
$or, 13p = 130$
$\therefore p = 10$

Creative Section

5) Find the arithmetic mean (or average) from the data given in the following tables.

a) 
X 0-10 10-20 20-30 30-40 40-50
f 5 7 8 6 4

Solution:
Arranging given data in a table,
X mid-value (m) f fm
0-10
10-20
20-30
30-40
40-50 		5
15
25
35
45 		5
7
8
6
4 		25
105
200
210
180
Total
N = 30 $\sum$fm= 720

Now,
Arithmetic mean ($\overline{x}$) = $\dfrac{\sum fm}{N}$

$= \dfrac{720}{30}$

$= 24$


c)
Wages (Rs) 50-60 60-70 70-80 80-90 90-100
No. of workers 6 8 12 8 6

Solution:
X mid-value (m) f fm
50-60
60-70
70-80
80-90
90-100 		55
65
75
85
95
 		6
8
12
8
6 		330
520
900
680
570
Total N=40 $\sum$fm=3000

Now,
Arithmetic mean ($\overline{x}$) = $\dfrac{\sum fm}{N}$

$= \dfrac{3000}{40}$

$= 75$
e)
Marks 20-30 30-40 40-50 50-60 60-70 70-80
No. of students 4 5 2 4 3 2

Solution
X mid-value (m) f fm
20-30
30-40
40-50
50-60
60-70
70-80 		25
35
45
55
65
75		 4
5
2
4
3
2		 100
175
90
220
195
150
Total N=20 $\sum$fm=930

Now,
Arithmetic mean ($\overline{x}$) = $\dfrac{ \sum fm}{N}$

$= \dfrac{930}{20}$

$= 46.5$


About vedanta EXCEL in MATHEMATICS Book 10

Author: Hukum Pd. Dahal
Editor: Tara Bahadur Magar

Vanasthali, Kathmandu, Nepal
+977-10-4382404, 01-4362082
vedantapublication@gmail.com

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Exercise 17.1 | Statistics - Mean | vedanta Excel in Mathematics | Class 10 is a collection of the solutions related to exercises of mean of grouped and continuous data from Statistics Chapter for Nepal's Secondary Education Examination (SEE) appearing students.

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