Mean
Arithmetic Means
If the total sum observation is divided by a total number of observations, then it is called arithmetic mean. It is denoted by \(\overline{X}\) (Read as Xbar)
∴ Arithmetic Mean = \(\frac{Total\;sum\;of\;observation}{Total\;no.\;of\;observation}\)
For example,
Arithmetic mean of 1, 3, 7, 11, & 13
= \(\frac{1+3+7+11+13}{5}\)
= \(\frac{35}{5}\)
= 7
 Calculation of Mean for individual series
The mean of individual series is calculated by adding all the observation and dividing the sum by the total number of observation.
If x_{1}, x_{2}, x_{3}, …………..x_{n} are be n variants value of variable a. Then arithmetic mean is denoted by
\(\overline{X}\) = \(\frac{x_1+ x_2+ x_3+ …………..x_n}{n}\) = \(\frac{∑X}{n}\)
Where ∑X = sum of n observation or items
n = no. of observations or items.
X = variable.  Calculation of Mean for discrete series
Mean for discrete series can be calculated by Mean \(\overline{X} = \frac{sum\;of\;the\;product\;of\;f\;and\;x}{sum\;of\;f}\)
= \(\frac{∑fx}{N}\)  Calculation of Mean for continuous series
For calculating mean in continuous series the following formulae is used:
\(\overline{X}\) = \(\frac{∑fx}{N}\) where, F = Frequency and m = midvalue.
 If the total sum observation is divided by total number of observations, then it is called arithmetic mean. It is denoted by \(\overline{X}\).
 The mean of individual series is calculated by adding all the observation and dividing the sum by the total number of observation.
Solution:
Now,
Sum of a numbers = 12 + 15 + 18 + 19 + 15 + 17 + 16 + 10 + 13 + 15
= 150
number = 10
We know that,
or, Mean (\(\overline{X}\)) = \(\frac{\sum{x}}{n}\)
or,Mean (\(\overline{X}\)) = \(\frac{150}{10}\)
\(\therefore\) Mean (\(\overline{X}\)) = 15
Solution:
Sum of given terms (\(\sum\)x) = 66 mph + 57 mph + 71 mph + 54 mph + 69 mph + 58 mph = 375
numbers of terms (n) = 6
We have,
Mean (\(\overline{X}\)) = \(\frac{\sum{x}}{n}\)
= \(\frac{375}{6}\)
= 62.5
The mean driving speed is 62.5 mph.
Solution:
Sum of given terms (\(\sum\)x) = 89+ 73+ 84+ 91+ 87+ 77+ 94 = 515
numbers of terms (n) =7
We have,
Mean (\(\overline{X}\)) = \(\frac{\sum{x}}{n}\)
= \(\frac{515}{7}\)
= 73.57
Hence, The mean test score is 73.57
Solution:
Sum of given terms (\(\sum\)x) = $1.79+ $1.61+ $1.96+ $2.08 = $7.44
numbers of terms (n) = 4
We have,
Mean (\(\overline{X}\)) = \(\frac{\sum{x}}{n}\)
= \(\frac{7.44}{7}\)
= 1.86
Hence, The mean gasoline price is $1.86
Solution:
Now,
Sum of numbers = 3 + 7 + 10 + 15 + x
= 35 + x
We know that,
or, Mean (\(\overline{X}\)) = \(\frac{\sum{x}}{n}\)
or, \(\frac{12}{1}\)= \(\frac{35 + x}{n}\)
or, 60 = 35 + x
or, x = 60  35
or, x = 25
\(\therefore\) The value of x is 25.
Solution:
Sum of given terms (\(\sum\)x) = 2.6 min+ 7.2 min+ 3.5 min+ 9.8 min+ 2.5 min =25.6
numbers of terms (n) =5
We have,
Mean (\(\overline{X}\)) = \(\frac{\sum{x}}{n}\)
= \(\frac{25.6}{5}\)
= 5.12
Hence, The mean swimming time to the nearest tenth is 5.12 min.
Solution:
Sum of given terms (\(\sum\)x) =
2.7 hr + 8.3 hr + 3.5 hr + 5.1 hr + 4.9 hr = 24.5hr
numbers of terms (n) =5
We have,
Mean (\(\overline{X}\)) = \(\frac{\sum{x}}{n}\)
= \(\frac{24.5}{5}\)
= 4.9
Hence, The mean race time is 4.9 hr
Solution:
Now,
Sum of number =m + m+2 + m+4 + m+6 + m+8
= 5m + 20
We know that,
or, Mean (\(\overline{X}\)) = \(\frac{\sum{x}}{n}\)
or, \(\frac{13}{1}\) = \(\frac{5m + 20}{5}\)
or, 65 = 5m + 20
or, 5m = 65  20
or, 5m = 45
or, m = \(\frac{45}{5}\)
or, m = 9
\(\therefore\) The value of m is 9.
Solution:
Now,
Sum of a numbers =y + y+3 + y+7 + y+10 + y+13 + y+15
= 6y + 48
We know that,
or, Mean (\(\overline{X}\)) = \(\frac{\sum{x}}{n}\)
or, 18 = \(\frac{6y + 48}{6}\)
or, 108 = 6y + 48
or, 108  48 = 6y
or, 6y = 60
or, y = \(\frac{60}{6}\)
or, y = 10
\(\therefore\) The value of y is 10.
Find the mean of the following data.
x  5  10  15  20  25 
f  2  6  10  5  2 
Solution:
x  f  fx 
5  2  10 
10  6  60 
15  10  150 
20  5  100 
25  2  50 
Total  25  370 
Now,
or, \(\overline{X}\) = \(\frac{\sum{x}}{n}\)
or, \(\overline{X}\) = \(\frac{370}{25}\)
\(\therefore\) \(\overline{X}\) = 14.8
Find the mean of the following continuous data.
Class Interval  0 6  6 12  12  18  18  24  24  30 
Frequency  2  4  10  6  2 
Solution:
Class Interval  frequency  MidValue  F.x 
0  6  32  3  96 
6  12  4  9  36 
12  18  10  15  150 
18  24  6  21  126 
24  30  2  27  54 
N = 24  \(\sum{fx}\) = 372 
We know that,
or, \(\overline{X}\) = \(\frac{\sum{fx}}{N}\)
or, \(\overline{X}\) = \(\frac{372}{24}\)
\(\therefore\)\(\overline{X}\) =15.5
Find the mean of the following data.
Weight (in kg)  20  24  30  32  35 
No. of students  4  6  8  5  2 
Solution:
Weight  No. of students  fx 
20  4  80 
24  6  144 
30  8  240 
32  5  160 
35  3  70 
N = 26  \(\sum{fx}\) = 694 
We know that,
or, \(\overline{X}\) = \(\frac{\sum{fx}}{N}\)
or, \(\overline{X}\) = \(\frac{694}{26}\)
\(\therefore\) \(\overline{X}\) = 26.69 kg
Find the mean of the following continuous data.
marks obtained 
10 20 
20  30 
30  40 
40  50 
50  60 
No. of students 
4 
6 
10 
3 
2 
Solution:
Marks  Students  midvalue  fx 
10  20  4  15  60 
20  30  6  25  150 
30  40  10  35  350 
40  50  3  45  135 
50  60  2  55  110 
N = 25  \(\sum{fx}\) = 805 
We know that,
or, \(\overline{X}\) = \(\frac{\sum{fx}}{N}\)
or, \(\overline{X}\) = \(\frac{372}{25}\)
\(\therefore\) \(\overline{X}\) =14.88
Find the mean of the following data.
Height (in cm) 
58 
60 
62 
64 
66 
68 
No. of plants 
12 
14 
20 
13 
8 
5 
Solution:
Height  No. of plants  fx 
58  12  696 
60  14  840 
62  20  1240 
64  13  832 
66  8  528 
68  5  340 
N = 72  \(\sum{fx}\) = 4478 
We know that,
or, \(\overline{X}\) = \(\frac{\sum{fx}}{N}\)
or, \(\overline{X}\) = \(\frac{4478}{72}\)
\(\therefore\) \(\overline{X}\) = 62.19 cm
Find the mean of the following continuous data.
Marks 
5  10 
10  15 
15  20 
20  25 
25  30 
No. of students 
2 
4 
8 
6 
4 
Solution:
Marks  Students  MidValue  fx 
5  10  2  7.5  15 
10  15  4  12.5  50 
15  20  8  17.5  140 
20  25  6  22.5  135 
25  30  4  27.5  110 
N = 24  \(\sum{fx}\) = 450 
We know that,
or, \(\overline{X}\) = \(\frac{\sum{fx}}{N}\)
or, \(\overline{X}\) = \(\frac{450}{24}\)
\(\therefore\) \(\overline{X}\) = 18.75
The mean of the given data is 21. Find the value of p.
class interval  0 8  8  16  16  24  24  32  32  40 
frequency  5  9  10  p  8 
Solution:
class interval  x  f  fx 
0  8  4  5  20 
8  16  12  9  108 
16  24  10  10  200 
24  32  28  p  28 + p 
32  40  36  8  288 
N = 32 + p  \(\sum{fx}\) = 616 + 28p 
we know that,
or, \(\overline{X}\) = \(\frac{\sum{fx}}{N}\)
or, 21= \(\frac{616 + 28p}{32 + p}\)
or, 672 + 21p = 616 + 28p
or, 672  616 = 28p  21p
or, 56 = 7p
or, p = \(\frac{56}{7}\)
or, p = 8
\(\therefore\) The required mean is 8.

Marie has the following data:
2, 2, k, 14, 3
If the mean is 6, which number could k be?
9
5
6
1

Find mean.
9, 5, 7, 0, 7, 4, 4, 1, 7, 6
6
7
5
2

Find mean.
4, 8, 12, 16 , 20
8
16
12
5

Find mean.
6, 16, 20, 30, 36
20
21.6
25.02
46

What is the mean?
7, 0, 6, 6, 5, 6, 3, 1, 6, 8
3
6
2
1

Brinda tallied the number of times her favourite new song played on 8 radio stations this week. She tallied:
4 plays 5 plays 6 plays 8 plays 4 plays 7 plays 9 plays 5 plays What was the mean number of plays?
4
5
9
6

Find the value of k from the given data whose mean is 23.
10, 17, 19, k, 27 , 29, 33
41
25
12
26

Find the value of x from the given data whose mean is 8.
x, 7, 9, 8, 10, 12
12
56
2
4

Find the value of P from the given data whose mean is 34.
20, 24, 32, 40, P & 52
36
25
45
21

Find the value of x from the given data whose mean is 22.
x, x+4, x+8, x+12, x+16
25
14
15
20

Find the value of Q if the mean of 10, 22, 12, 8, 14, Q and 20 is 16.
20
18
12
26

Find the mean.
90, 50, 70, 30, 20, 10
80
60
45
55

The age (in years) of 15 students are as follows. Calculate their average age.
12, 13, 11, 12, 15 , 14 , 11, 12 , 12, 11, 13, 12, 11
12
12.33
14
13

The mean of 5, 7, 9, x and 15 is 10, find x.
15
16
12
14

The mean of 3, 7, 10, 15 and x is 12, find x
20
15
25
5

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