∑
Statistics Mean (Individual)
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Mathematics: Mean for Individual Data
Standard Formula
For individual series, Arithmetic Mean (x̄) is:
x̄ =
∑ x
N
Mean =
Sum of all observations (∑x)
Total number of observations (N or ∑f)
Worked Examples & Practice Questions
Question No. 1
Find the Mean (x̄) of the dataset: 12, 15, 18, 22, 25, 30
Solution:
Given dataset (x) = 12, 15, 18, 22, 25, 30
Number of terms (N) = 6
∑x = 12 + 15 + 18 + 22 + 25 + 30 = 122
Using formula,
x̄ = ∑xN
= 1226
∴ x̄ = 20.33
Question No. 2
Calculate the mean of the test scores: 45, 50, 62, 38, 70, 55, 40
Solution:
Number of scores (N) = 7
∑x = 45 + 50 + 62 + 38 + 70 + 55 + 40 = 360
Now,
Mean (x̄) = ∑xN
= 3607
∴ Mean = 51.43 marks
Question No. 3
Find the average (Mean) of the first 10 counting numbers.
Solution:
First 10 counting numbers = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10
N = 10
∑x = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 = 55
x̄ = 5510
∴ x̄ = 5.5
Question No. 4
If the mean of 5 numbers is 20 and four of the numbers are 15, 18, 22, and 24, find the missing fifth number.
Solution:
Let the missing 5th number = k
Given, Mean (x̄) = 20, N = 5
∑x = 15 + 18 + 22 + 24 + k = 79 + k
x̄ = ∑xN
⇒ 20 = 79 + k5
or, 100 = 79 + k
or, k = 100 – 79
∴ k = 21
Question No. 5
Ram scored 75, 80, 65, and 70 marks in four subjects. If his average score across 5 subjects is 74, how many marks did he score in Social Studies?
Solution:
Let marks in Social Studies = M
N = 5, Mean (x̄) = 74
Sum of 4 subjects = 75 + 80 + 65 + 70 = 290
∑x = 290 + M
74 = 290 + M5
or, 370 = 290 + M
or, M = 370 – 290 = 80
∴ Marks in Social Studies = 80
Question No. 6
In a cricket match, a bowler conceded 4, 8, 2, 12, 6, and 4 runs in 6 overs. Find his mean economy rate.
Solution:
Total Overs (N) = 6
∑x = 4 + 8 + 2 + 12 + 6 + 4 = 36
Mean Economy Rate = 366
∴ Mean Economy Rate = 6 runs/over
Question No. 7
A team kept an average run rate of 6 runs/over for the first 15 overs. What mean run rate must they maintain in the last 5 overs to target 150 total runs in 20 overs?
Solution:
Runs in 15 overs = 15 × 6 = 90 runs
Target Total Runs = 150 runs
Runs required in last 5 overs = 150 – 90 = 60
Required Mean = 605
∴ Required Run Rate = 12 runs/over
★ IMPORTANT QUESTION ★
b) ∑x = 255, N = 8 x̄ =
c) Let 9th match runs = R Target Mean = 40, N’ = 9 40 =
Question No. 8
Rohit Paudel’s run scores in eight matches: 62, 2, 62, 16, 16, 96, 1, 2.
a) Write the formula to find mean.
b) Find average mean run score in 8 matches.
c) How many runs are needed in the 9th match to make the average 40?
Solution:
a) Mean (x̄) = ∑xN
b) ∑x = 255, N = 8 x̄ =
2558
= 31.875 runs
c) Let 9th match runs = R Target Mean = 40, N’ = 9 40 =
255 + R9
360 = 255 + R ⇒ R = 105
∴ Required Runs = 105
★ IMPORTANT QUESTION ★
Question No. 9
If the mean of 4, 10, 3, 6, a, 9, 10 is 7, find the value of a.
Solution:
Given, N = 7, x̄ = 7
∑x = 4 + 10 + 3 + 6 + a + 9 + 10 = 42 + a
7 = 42 + a7
49 = 42 + a ⇒ a = 7
∴ Value of a = 7
★ EXTRA CREATED QUESTION ★
Question No. 10
Recorded temperatures over 6 days: 28°C, 32°C, 30°C, 35°C, p°C, 29°C. If mean temperature is 31°C, find p.
Solution:
N = 6, x̄ = 31°C
∑x = 154 + p
31 = 154 + p6
186 = 154 + p ⇒ p = 32°C
∴ Missing temperature = 32°C
Question No. 11
If ∑f = 5 and ∑x = 65, find the value of Mean (x̄).
Solution:
∑x = 65, N = ∑f = 5
x̄ = 655
= 13
∴ x̄ = 13
Question No. 12
Given ∑x = 240 and N = 8, calculate the value of x̄.
Solution:
∑x = 240, N = 8
x̄ = 2408
= 30
∴ x̄ = 30
Question No. 13
If ∑x = m + 77, N = 10, and Mean (x̄) = 8, find m.
Solution:
8 = m + 7710
80 = m + 77 ⇒ m = 3
∴ m = 3
Question No. 14
Given ∑x = 2m + 10 and ∑f = m + 5, calculate Mean (x̄).
Solution:
|
x̄ = 2m + 10m + 5
|
(Given equation for Arithmetic Mean) |
|
= 2(m + 5)m + 5
|
(Factorise numerator by taking out common factor 2) |
|
= 2 × m + 5m + 5
|
(Separate the constant factor 2) |
| = 2 × 1 | (Cancel common binomial term (m + 5) from top and bottom) |
| = 2 | (Final simplified result) |
Question No. 15
Calculate the mean expenditure of Rai’s family:
| Month | Baishakh | Jestha | Ashadh | Shrawan |
|---|---|---|---|---|
| Expenditure | 35,000 | 40,000 | 60,000 | 45,000 |
Solution:
N = 4
∑x = 35,000 + 40,000 + 60,000 + 45,000 = 1,80,000
x̄ = 1,80,0004
∴ Mean Expenditure = Rs. 45,000
Question No. 16
Average monthly expenditure of the family:
| Month | Shrawan | Bhadra | Ashwin | Kartik | Mangsir |
|---|---|---|---|---|---|
| Expenditure | 30,000 | 25,000 | 50,000 | 45,000 | 30,000 |
Solution:
N = 5
∑x = 1,80,000
x̄ = 1,80,0005
∴ Average Expenditure = Rs. 36,000
Out of Topic (Combined Mean)
Question No. 17
30 boys mean weight = 45 kg, 20 girls mean weight = 40 kg. Find combined mean weight.
Solution:
Boys: N1 = 30, x̄1 = 45
Girls: N2 = 20, x̄2 = 40
x̄12 = (30×45) + (20×40)30 + 20
x̄12 = 215050
= 43 kg
∴ Combined Mean = 43 kg
Question No. 18
Average salary of 50 males = Rs. 25,000 and 40 females = Rs. 22,000. Calculate combined mean.
Solution:
Males: N1 = 50, x̄1 = 25000
Females: N2 = 40, x̄2 = 22000
x̄12 = (50×25000) + (40×22000)50 + 40
x̄12 = 213000090
∴ Combined Mean = Rs. 23,666.67
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