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, \sumfm = 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, \sumfm = 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 \sumfm= 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 \sumfm=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 \sumfm=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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