Arithmetic Mean

Subject: Compulsory Mathematics

Find Your Query

Overview

The arithmetic mean is defined as the total sum of observations divided by the total number of observations. Here we study about the arithmetic mean of continuous series. Calculation of arithmetic means in continuous series When the number of items is large, we have to divide them into groups. Such groups are known as simple classes. It should be noted that there is no gap between any two successive intervals. So the data is continuous and the series is called continuous series.
Arithmetic Mean

The word "statistics " refers two meanings. In the singular sense, it deals with the collection, presentation, analysis and interpretation of numerical data and helps in making a decision. In the plural sense, it refers to the numerical facts and figures are sometimes known as statistical data.

Measures of central tendency

.

A typical value which represents the characteristics of the entire mass of huge data is called the central value of the whole distribution. A measure of the central tendency is also known as a measure of location or an average.
The various measure of central tendency are as follow:

  1. Arithmetic Mean (average)
  2. Median
  3. Mode

Arithmetic Mean

The arithmetic mean is defined as the total sum of observations divided by the total number of observations. Here we study about the arithmetic mean of continuous series.

Scholarships after +2 Abroad Studies Opportunities

Calculation of arithmetic means in continuous series

When the number of items is large, we have to divide them into groups. Such groups are known as simple classes. It should be noted that there is no gap between any two successive intervals. So the data is continuous and the series is called continuous series.

Arithmetic Mean can be calculated in three different methods.

  1. \(Direct \: method, \overline{X} = \frac{\sum fm}{n}\), where m is the mid-value of class intervals.
  2. \(Short \: cut \: method,\overline{X} = A + \frac{\sum fd}{n} \), where d= X - A and A is assumed mean.
  3. \(Step \: deviation \: method, \overline{X} = A + \frac{\sum fd'}{n}\) \( \times h \), where d' = \(\frac{d}{h}\) and h is the length of class interval.


    .

Things to remember

Arithmetic Mean can be calculated in three different methods.

  1. \(Direct \: method, \overline{X} = \frac{\sum fm}{n}\), where m is the mid-value of class intervals.
  2. \(Short \: cut \: method,\overline{X} = A + \frac{\sum fd}{n} \), where d= X - A and A is assumed mean.
  3. \(Step \: deviation \: method, \overline{X} = A + \frac{\sum fd'}{n}\) \( \times h \), where d' = \(\frac{d}{h}\) and h is the length of class interval.
  • It includes every relationship which established among the people.
  • There can be more than one community in a society. Community smaller than society.
  • It is a network of social relationships which cannot see or touched.
  • common interests and common objectives are not necessary for society.
Videos for Arithmetic Mean
Central Tendency - Mean Median Mode Range - MathHelp.com
Combined Arithmetic Mean
Statistics Arithmetic Mean Type 1
Statistics Arithmetic Mean Type 2
Questions and Answers

Solution:

Mean(\(\overline X\))= 5o
\(\sum fx = 750 \)
N = ?

\begin{align*} Mean\:(\overline {X}) &= \frac{\sum fx}{N}\\ 50 &= \frac{750}{N}\\ or, N &= \frac{750}{50}\\ \therefore N &= 15 \: \: \: \: _{Ans} \end{align*}

Solution:

\(Mean \: (\overline{X}) =60 \\ \sum fx = 960 \\ Number \: of \: term \: (N) = ? \)

\begin{align*} Mean (\overline {X}) &= \frac{\sum fx}{N}\\ 60 &= \frac{960}{N}\\ or, N &= \frac{960}{60}\\ \therefore N &= 16 \: _{Ans} \end{align*}

Solution:

\(Mean \: (\overline{X}) =12 \\ \sum fx = 70 +10a\\ No. \: of \: term \: (N) = 5+a \)

\begin{align*}Mean \: (\overline{X}) &= \frac{\sum fx}{N}\\ 12 &= \frac{70 + 10a}{5 + a}\\ or, 60 +12a &=70 + 10a\\ or,12a -10a &= 70 - 60\\ or, 2a &= 10 \\ or, a &= \frac{10}{2}\\ \therefore a &= 5 \: \: _{Ans} \end{align*}

Solution:

\(Mean \: (\overline{X}) = 13 \\ No. \: of \: terms \: (N) = 5 + a \)

\begin{align*} Mean (\overline{X}) &= \frac{\sum X}{N} \\ or, 13 &= \frac{4 + 8+12+x+25}{5}\\ or, 65 &=49 + x\\ or, x &= 65 - 49 \\ x &= 16 \:\: _{Ans} \end{align*}

Solution:

First item mean \( (\overline {X_{1}}) = 7 \)
Number of first item \(n_{1} = 4 \)
Second item mean \( (\overline {X_{2}}) = 12\)
No. of Second item \(n_{2}\)= 3
Mean of 7 items \(\overline{X}= ?\)

\begin{align*} ( \overline {X}) &= \frac{n_1 \overline{X}_1 + n_2 \overline{X}_2}{n_1+n_2}\\ &= \frac{4 \times 7 + 3 \times 12}{4 + 3}\\ &= \frac{28 + 36}{7}\\ &= \frac{64}{7}\\ &= 9.14 \: \: \: _{ans} \end{align*}

\(Mean (\overline{X}) = 40 \\ No. \: of \: terms (N)= 7 \\ k = ? \)

\begin{align*} Mean(\overline{X}) &= \frac{\sum X}{N} \\ or, 40 &= \frac{10+20+30+40+50+ 60+30+k}{7}\\ or, 280 &= 240 + k\\ or, k &= 280 - 240\\ \therefore k &= 40 \: \: \: \: \: \: \: \: \: _{Ans} \end{align*}

Solution:

Expenditure (in Rs.)x Frequency(f) fx
24 2 48
25 4 100
30 3 90
35 4 140
40 2 80

\begin{align*} Mean(\overline{X}) &= \frac{\sum fx}{N}\\ &= \frac{458}{15}\\ &= 30.53 \\ \therefore Average \: expenditure &= Rs \: 30.53 \end{align*}

Solution:

Let, the age of remaining students be x.

\(average \: age (\overline{X}) = 9 \\ Number \: of \: terms (N) = 5 \)

\begin{align*} (\overline{X}) &= \frac{\sum X}{N}\\ or, 9 &= \frac{5+7+8+15+x}{5}\\ or, 45 &= 35 + x \\ \therefore x &=45 -35 = 10 \\ \: \\ \therefore & \text{The age of remaining student is 10} \end{align*}

Solution:

Calculation of mean

Class interval mid-value (m) frequency (f) fm
0-10 5 3 15
10-20 15 5 75
20-30 25 6 150
30-40 35 5 175
40-50 45 2 90
50-60 55 4 220
N = 25

We know that,

\begin{align*} Mean (\overline{X} ) &= \frac{\sum fm}{N}\\ &= \frac{725}{25}\\ &= 29 \: _{Ans} \end{align*}

Solution:

Calculation of mean

Class interval cf. f mid-value (m) fm
10-20 4 4 15 60
20-30 16 12 25 300
30-40 56 40 35 1400
40-50 97 41 45 1845
50-60 124 27 55 1485
60-70 137 13 65 845
70-80 146 9 75 675
80-90 150 4 85 340
N = 150 \(\sum fm=6950\)

\begin{align*}Mean \: (\overline{X})&= \frac{\sum fm}{N}\\ &= \frac{6950}{150}\\ &= 46.33 \: _{Ans} \end{align*}

Solution:

Calculation of mean

Wages cf f m fm
50-60 6 6 55 330
60-70 14 8 65 520
70-80 26 12 75 900
80-90 34 8 85 680
90-100 40 6 95 570
N=40 \(\sum fm= 3000\)

\begin{align*} Mean (\overline {X})&= \frac{\sum fm}{N}\\ &= \frac{3000}{40}\\ &= 75 \: _{Ans} \end{align*}

Solution:

Calculation of mean

Wages cf f m fm
50-60 6 6 55 330
60-70 14 8 65 520
70-80 26 12 75 900
80-90 34 8 85 680
90-100 40 6 95 570
N=40 \(\sum fm= 3000\)

\begin{align*} Mean (\overline {X})&= \frac{\sum fm}{N}\\ &= \frac{3000}{40}\\ &= 75 \: _{Ans} \end{align*}

Solution:

Calculation of mean

class interval midvalue f fm
0-10 5 5 25
10-20 15 7 105
20-30 25 8 200
30-40 35 4 140
40-50 45 6 270
N= 30 \( \sum fm = 740 \)

\begin{align*} Mean (\overline{X}) &= \frac{\sum fm}{N}\\ &= \frac{740}{30}\\ &= 24.6 \: _{ans} \end{align*}

Solution:

Calculation of mean

Marks f m fm
10-20 2 15 30
20-30 5 25 125
30-40 7 35 245
40-50 6 45 270
50-60 3 55 165
60-70 2 65 130
N=25 \(\sum fm = 965\)

We know that,

\begin{align*} Mean (\overline {X}) &= \frac{\sum fm }{N} \\ &= \frac{965}{25}\\ &= 38.6 \: \: _{Ans} \end{align*}

Solution:

To find the value of p.

X f m fm
5-15 5 10 50
15-25 8 20 160
25-35 p 30 30p
35-45 9 40 360
45-55 7 50 350
55-65 1 60 60
N = p + 30 \( \sum fm = 30p+ 980 \)

We know that,

\begin{align*}Mean (\overline{X})&= \frac{\sum fm}{N}\\ 32 &= \frac{30p + 980}{p + 30}\\ or, 32p + 960 &= 30p + 980\\ or, 32p -30p &= 980 - 960 \\ or, p &= \frac{20}{2}\\ \therefore p&=10 \: _{Ans} \end{align*}

Solution:

To find the value of p.

Solution:

To find the value of p.

Age in year No. of teachers (f) m fm
10-20 3 15 45
20-30 8 25 200
30-40 15 35 525
40-50 p 45 45p
50-60 4 55 220
N=p+30 \( \sum fm = 45p + 990\)

We know that,

\begin{align*} Mean \: (\overline {X}) &= \frac{\sum fm}{N}\\ 36 &= \frac{45p + 990}{p+30}\\ or, 1080 + 36p &= 45p +990\\ or, 45p - 36p &= 1080 -990 \\ or, 9p &= 90 \\ \therefore p &= \frac{90}{9} = 10 \: _{Ans} \end{align*}

Solution:

To find the value of p.

Solution:

To find the value of p.

Age in year No. of teachers (f) m fm
10-20 3 15 45
20-30 8 25 200
30-40 15 35 525
40-50 p 45 45p
50-60 4 55 220
N=p+30 \( \sum fm = 45p + 990\)

We know that,

\begin{align*} Mean \: (\overline {X}) &= \frac{\sum fm}{N}\\ 36 &= \frac{45p + 990}{p+30}\\ or, 1080 + 36p &= 45p +990\\ or, 45p - 36p &= 1080 -990 \\ or, 9p &= 90 \\ \therefore p &= \frac{90}{9} = 10 \: _{Ans} \end{align*}

Solution:

To find the value of m.

Wages f m fm
15-25 4 20 80
25-35 6 30 180
35-45 12 40 480
45-55 m 50 50m
55-65 3 60 180
N = m + 25 \( \sum fm = 50m + 920\)

\begin{align*} Mean (\overline{X}) &= \frac{\sum fm}{N}\\ 39 &= \frac{50m+920}{m+25}\\ or, 39m + 975 &= 50m + 920 \\ or, 50m - 39m &= 975 - 920 \\ or, 11m &= 55 \\ or, x&=\frac{55}{11}\\ \therefore m &= 5 \: _{Ans} \end{align*}

Solution:

To find the value of k.

Class f m fm
0-20 15 10 150
20-40 k 30 30k
40-60 21 50 1050
60-80 29 70 2030
80-100 17 90 1530
N=82+k \( \sum fm = 30k + 4760 \)

\begin{align*} Mean \: (\overline{X} ) &= \frac{\sum fm}{N}\\ 53 &= \frac{30k + 4760}{82 + k}\\ or, 4346 + 53k &= 30k + 4760 \\ or, 53k - 30k &= 4760 - 4346 \\ or, 23k &= 414 \\ or, k &= \frac{414}{23}\\ \therefore k &= 18 \: _{Ans} \end{align*}

Solution:

Calculating mean

Marks frequency (f) mid value (m) fm
20-30 12 25 300
30-40 7 35 245
40-50 8 45 360
50-60 3 55 165
60-70 10 65 650
N = 40 \(\sum fm = 1720\)

\begin{align*} Mean \: (\overline{X}) &= \frac{\sum fm}{N}\\ &= \frac{1720}{40}\\ &= 43 \: _{Ans} \end{align*}

Solution:

To find the value of x and y

Marks f mid value fm
0-10 4 5 20
10-20 x 15 15x
20-30 10 25 250
30-40 y 35 35y
40-50 6 45 270
50-60 4 55 220
\( N= x+y +24\) \(\sum fm= 15x +35y +760 \)

\begin{align*} Mean (\overline{X}) &= \frac{\sum fm}{N}\\ 31 &= \frac{15x +35y+760}{50}\\ or, 1150 - 760 &= 15x+35y \\ or, 15x + 35y &= 790 \: ... .. .. . .. (1) \\ x+y+24 &= 50 \\ or, y&=50-24-x \\ y&=26-x ....................(2) \\ Putting \: valu&e \: of \: y \: in\:equation \: ....(1)\\ 15x + 35y&=790\\ or, 15x + 35(26-x) &= 790 \\ or, 15x+910-35x &= 790\\ or, -20x &= 790-910 \\ or, x &= \frac{-120}{-20}\\ \therefore x &= 6 \\ \: \\ putting \: value \: of \: x & \: in \: equation .. (2) \\ y &= 26-6 \\ \therefore y&=20 \end{align*}

Quiz

© 2019-20 Kullabs. All Rights Reserved.