Note on Mean

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Arithmetic Means

Arithmetic Mean
Arithmetic Mean

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%
  • Marie has the following data:

     2, 2, k, 14, 3 

    If the mean is 6, which number could k be?

    1
    6
    9
    5
  • Find mean.

    -9, -5, -7, 0, -7, -4, -4, -1, -7, -6

    -5
    -7
    -6
    2
  • Find mean.

    4, 8, 12, 16 , 20

    16
    5
    8
    12
  • Find mean.

    6, 16, 20, 30, 36

    46
    21.6
    25.02
    20
  • What is the mean?

    7, 0, 6, 6, -5, 6, 3, -1, 6, -8

    -1
    2
    -6
    3
  • 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?

    5
    4
    6
    9
  • Find the value of k from the given data whose mean is 23.

    10, 17, 19, k, 27 , 29, 33

    25
    12
    26
    41
  • Find the value of x from the given data whose mean is 8.

    x, 7, 9, 8, 10, 12

    56
    4
    2
    12
  • Find the value of P from the given data whose mean is 34.

    20, 24, 32, 40, P & 52

    36
    45
    21
    25
  • Find the value of x from the given data whose mean is 22.

    x, x+4, x+8, x+12, x+16

    25
    15
    20
    14
  • Find the value of Q if  the mean of 10, 22, 12, 8, 14, Q and 20 is 16.

     

    12
    26
    18
    20
  • Find the mean.

    90, 50, 70, 30, 20, 10

    45
    55
    60
    80
  • 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

    12.33
    14
    12
    13
  • The mean of 5, 7, 9, x and 15 is 10, find x. 

    12
    15
    16
    14
  • The mean of 3, 7, 10, 15 and x is 12, find x

    25
    20
    5
    15
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