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:
 Arithmetic Mean (average)
 Median
 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.
 \(Direct \: method, \overline{X} = \frac{\sum fm}{n}\), where m is the midvalue of class intervals.
 \(Short \: cut \: method,\overline{X} = A + \frac{\sum fd}{n} \), where d= X  A and A is assumed mean.
 \(Step \: deviation \: method, \overline{X} = A + \frac{\sum fd'}{n}\) \( \times h \), where d' = \(\frac{d}{h}\) and h is the length of class interval.
Arithmetic Mean can be calculated in three different methods.
 \(Direct \: method, \overline{X} = \frac{\sum fm}{n}\), where m is the midvalue of class intervals.
 \(Short \: cut \: method,\overline{X} = A + \frac{\sum fd}{n} \), where d= X  A and A is assumed mean.
 \(Step \: deviation \: method, \overline{X} = A + \frac{\sum fd'}{n}\) \( \times h \), 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*}
Find the mean from the following data.
Classinterval  1020  1030  1040  1050  1060  1070  1080  1090 
Frequency  4  16  56  97  124  137  146  150 
Solution:
Calculation of mean
Class interval  cf.  f  midvalue (m)  fm 
1020  4  4  15  60 
2030  16  12  25  300 
3040  56  40  35  1400 
4050  97  41  45  1845 
5060  124  27  55  1485 
6070  137  13  65  845 
7080  146  9  75  675 
8090  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*}
Calculate the mean from the following data.
X  010  1020  2030  3040  4050 
f  5  7  8  4  6 
Solution:
Calculation of mean
class interval  midvalue  f  fm 
010  5  5  25 
1020  15  7  105 
2030  25  8  200 
3040  35  4  140 
4050  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 
1020  3  15  45 
2030  8  25  200 
3040  15  35  525 
4050  p  45  45p 
5060  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 
1020  3  15  45 
2030  8  25  200 
3040  15  35  525 
4050  p  45  45p 
5060  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  midvalue (m)  frequency (f)  fm 
010  5  3  15 
1020  15  5  75 
2030  25  6  150 
3040  35  5  175 
4050  45  2  90 
5060  55  4  220 
N = 25 
We know that,
\begin{align*} Mean (\overline{X} ) &= \frac{\sum fm}{N}\\ &= \frac{725}{25}\\ &= 29 \: _{Ans} \end{align*}
Calculate the average wages from the following data.
Wages  5060  5070  5080  5090  50100 
No. of people  6  14  26  34  40 
Solution:
Calculation of mean
Wages  cf  f  m  fm 
5060  6  6  55  330 
6070  14  8  65  520 
7080  26  12  75  900 
8090  34  8  85  680 
90100  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*}
Calculate the average wages from the following data.
Wages  5060  5070  5080  5090  50100 
No. of people  6  14  26  34  40 
Solution:
Calculation of mean
Wages  cf  f  m  fm 
5060  6  6  55  330 
6070  14  8  65  520 
7080  26  12  75  900 
8090  34  8  85  680 
90100  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*}
Compute the mean from the table given below.
Marks  1020  2030  3040  4050  5060  6070 
No. of students  2  5  7  6  3  2 
Solution:
Calculation of mean
Marks  f  m  fm 
1020  2  15  30 
2030  5  25  125 
3040  7  35  245 
4050  6  45  270 
5060  3  55  165 
6070  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*}
If the mean is 32, find the value of p from the following.
x  515  1525  2535  3545  4555  5565 
f  5  8  p  9  7  1 
Solution:
To find the value of p.
X  f  m  fm 
515  5  10  50 
1525  8  20  160 
2535  p  30  30p 
3545  9  40  360 
4555  7  50  350 
5565  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*}
If the mean of the data given below is 39. Calculate the value of m.
Wages  1525  2535  3545  4555  5565 
No. of workers  4  6  12  m  3 
Solution:
To find the value of m.
Wages  f  m  fm 
1525  4  20  80 
2535  6  30  180 
3545  12  40  480 
4555  m  50  50m 
5565  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*}
If the mean of the following data is 53, find the value k.
Class  020  2040  4060  6080  80100 
Frequency  15  k  21  29  17 
Solution:
To find the value of k.
Class  f  m  fm 
020  15  10  150 
2040  k  30  30k 
4060  21  50  1050 
6080  29  70  2030 
80100  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*}
Calculate mean from the following data.
Marks  2030  3040  4050  5060  6070 
Frequency  12  7  8  3  10 
Solution:
Calculating mean
Marks  frequency (f)  mid value (m)  fm 
2030  12  25  300 
3040  7  35  245 
4050  8  45  360 
5060  3  55  165 
6070  10  65  650 
N = 40  \(\sum fm = 1720\) 
\begin{align*} Mean \: (\overline{X}) &= \frac{\sum fm}{N}\\ &= \frac{1720}{40}\\ &= 43 \: _{Ans} \end{align*}
The mean and total number of students are 31 and 50 find the missing frequency.
Marks  010  1020  2030  3040  4050  5060 
No. of students  4  x  10  y  6  4 
Solution:
To find the value of x and y
Marks  f  mid value  fm 
010  4  5  20 
1020  x  15  15x 
2030  10  25  250 
3040  y  35  35y 
4050  6  45  270 
5060  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&=5024x \\ y&=26x ....................(2) \\ Putting \: valu&e \: of \: y \: in\:equation \: ....(1)\\ 15x + 35y&=790\\ or, 15x + 35(26x) &= 790 \\ or, 15x+91035x &= 790\\ or, 20x &= 790910 \\ or, x &= \frac{120}{20}\\ \therefore x &= 6 \\ \: \\ putting \: value \: of \: x & \: in \: equation .. (2) \\ y &= 266 \\ \therefore y&=20 \end{align*}

The mean of 50 observations was 250. It was detected on checking that the value of 165 was wrongly copied as 115 for computation of mean. Find the correct mean.
251
152
543
345

The mean of 40 observations was 160. It was detected on rechecking that the value of 165 was wrongly copied as 125 for computation of mean. Find the correct mean.
321
161
465
785

In a grouped data, the sum of the marks obtained by 50 students in mathematics is 3000.Find the average mark.
60
20
80
60

In a grouped data, the sum of the marks obtained by 50 students in mathematics is 3000.Find the average mark.
80
20
60
60

In a continuous series the value of a assumed mean is 35 and the sum of frequencies is 50.If mean of the data is 41 then find the sum of the product of deviation (d) and frequencies(f).
750
200
300
700

In a continuous series the value of a assumed mean is 25 and the sum of frequencies is 150.If mean of the data is 26.07 then find the sum of the product of deviation (d) and frequencies(f).
80
120
175
160

If the mean of a grouped data having ∑fm =306 is 18, find the value of N.
7
10
20
17

In a continuous frequency distribution data, the number of terms (N) = 40 and mean ((overline {X}) )= 75. If a class interval 4050 having frequency 10 is included in the series, find the new mean.
69
54
92
22

In a continuous frequency distribution data, the number of terms (N) = 40 and mean (overline {X}) )= 75. If a class interval 7080 having frequency 12 is excluded from the series, what will be the new mean?
75
40
55
85

In a continuous frequency distribution data, the number of terms (N) = 40 and mean ((overline {X}) )= 36. If a class interval 4050 having frequency 10 is excluded from the series, what will be the new mean?
65
33
45
21

In a continuous series the mean of 50 observation was 40. It was detected on checking that the class interval 3040 having frequency 10 was wrongly copied as class interval 3050 having frequency 10 for computation of mean.Find the correct mean.
69
59
49
39

In a continuous series the mean of 50 observation was 40. It was detected on checking that the class interval 3040 having frequency 10 was wrongly copied as class interval 3050 having frequency 10 for computation of mean.Find the correct mean.
69
59
39
49

In a continuous series the mean of 40 observation was 60. It was detected on checking that the class interval 2040 having frequency 10 was wrongly copied as class interval 2040 having frequency 10 for computation of mean.Find the correct mean.
40.25
35.30
61.25
31

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NivaIn the continuos series mean=15 a and £fx = 420 a find the number of terms(N) 
Feb 02, 2017 
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Ask any queries on this note.find the value of meanmark 010 020 030 040 050f 5 10 14 17 20 
Jan 29, 2017 
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