#### 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.

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.

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.

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 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

 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:

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:

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

 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

 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 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*}

0%

345

543

251

152

785

465

321

161

60

60

20

80

80

60

20

60

750

300

200

700

120

175

160

80

10

17

20

7

54

92

69

22

40
55
85
75

45

33

65

21

69

39

49

59

69

59

49

39

61.25

31

40.25

35.30