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
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 | 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
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 | 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
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 | 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
Marie has the following data:
2, 2, k, 14, 3
If the mean is 6, which number could k be?
Find mean.
-9, -5, -7, 0, -7, -4, -4, -1, -7, -6
Find mean.
4, 8, 12, 16 , 20
Find mean.
6, 16, 20, 30, 36
What is the mean?
7, 0, 6, 6, -5, 6, 3, -1, 6, -8
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?
Find the value of k from the given data whose mean is 23.
10, 17, 19, k, 27 , 29, 33
Find the value of x from the given data whose mean is 8.
x, 7, 9, 8, 10, 12
Find the value of P from the given data whose mean is 34.
20, 24, 32, 40, P & 52
Find the value of x from the given data whose mean is 22.
x, x+4, x+8, x+12, x+16
Find the value of Q if the mean of 10, 22, 12, 8, 14, Q and 20 is 16.
Find the mean.
90, 50, 70, 30, 20, 10
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
The mean of 5, 7, 9, x and 15 is 10, find x.
The mean of 3, 7, 10, 15 and x is 12, find x
No discussion on this note yet. Be first to comment on this note