## Note on Mean

• Note
• Things to remember
• Exercise
• Quiz

#### 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 X-bar)
∴ 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

1. 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 x1, x2, x3, …………..xn are be n variants value of variable a. Then arithmetic mean is denoted by $$\overline{X}$$
So, $$\overline{X}$$ = $$\frac{x_1+ x_2+ x_3+ …………..x_n}{n}$$ = $$\frac{∑X}{n}$$
Where ∑X = sum of n observation or items
n = no. of observations or items
X = variable

2. Calculation of Mean for discrete series
Mean for discrete series can be calculated by
$$\overline{X} = \frac{sum\;of\;the\;product\;of\;f\;and\;x}{sum\;of\;f}$$
= $$\frac{∑fx}{N}$$

3. Calculation of Mean for continuous series
For calculating mean in continuous series the following formulae is used:
$$\overline{X}$$ = A + $$\frac{∑fx}{N}$$ × i
where, F = Frequency and A = mid-value

• 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.
.

### Very Short Questions

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.

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

Solution:

 Class Interval frequency Mid-Value 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

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

Solution:

 Marks Students mid-value 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

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

Solution:

 Marks Students Mid-Value 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

0%

5
9
1
6

-6
-5
-7
2

8
12
5
16

25.02
46
21.6
20

-6
3
2
-1

5
9
4
6

41
25
26
12

56
4
2
12

45
21
36
25

20
14
25
15

26
20
12
18

60
80
55
45

12
13
14
12.33

15
16
12
14

20
15
25
5