Mean – Problems
1. Mean of 9 observations is 100 and mean of 6 observations is 80, then the mean of 15 observations is
Solutions: 9 observations is 100
6 observations is 80
n₁ = 9 and n₂ = 6
x̄₁ = 100 and x̄₂ = 80
We know that
\(\bar{x}=\frac{{{n}_{1}}{{{\bar{x}}}_{1}}+{{n}_{2}}{{{\bar{x}}}_{2}}}{m+n}\),
\(=\frac{9\times 100+6\times 80}{9+6}\),
\(=\frac{1380}{15}=92\).
2. Mean and made of the following data are respectively
Class |
fᵢ |
0 – 10 |
22 |
10 – 20 |
38 |
20 – 30 |
46 |
30 – 40 |
35 |
40 – 50 |
20 |
Solution:
Class |
xᵢ | fᵢ |
xᵢfᵢ |
0 – 10 |
5 | 22 | 110 |
10 – 20 | 15 | 38 |
570 |
20 – 30 |
25 | 46 | 1150 |
30 – 40 | 35 | 35 |
1225 |
40 – 50 |
42 | 20 | 900 |
161 |
3955 |
Mean \(=\frac{\sum{{{f}_{i}}{{x}_{i}}}}{\sum{{{f}_{i}}}}=\frac{3955}{161}\) = 24.56
Median = 24.26 (approximate)
Mean = 3 (median) – 2 (mean)
Mode = 3 (24.46) – 2 (24.46)
= 24.26.
3. Find the mean deviation about mean for the data in following table
Income per Year |
Number of Persons |
0 – 100 |
4 |
100 – 200 |
8 |
200 – 300 |
9 |
300 – 400 |
10 |
400 – 500 |
7 |
500 – 600 |
5 |
600 – 700 |
4 |
700 – 800 |
3 |
Solution:
Income per Year |
Number of Persons | Mid Value xᵢ | \({{d}_{i}}={{A}_{i}}+\frac{\sum{{{f}_{i}}{{d}_{i}}}}{\sum{{{f}_{i}}}}\times h\), A = 350 & h = 100 | fᵢdᵢ | |xᵢ – x̄| | fᵢ |xᵢ – x̄| |
0 – 100 | 4 | 50 | -3 | -12 | 308 |
1232 |
100 – 200 |
8 | 150 | -2 | -16 | 208 | 1664 |
200 – 300 | 9 | 250 | -1 | -9 | 108 |
972 |
300 – 400 |
10 | 350 | 0 | 0 | 8 | 80 |
400 – 500 | 7 | 450 | 1 | 7 | 92 |
644 |
500 – 600 |
5 | 550 | 2 | 10 | 192 | 960 |
600 – 700 | 4 | 650 | 3 | 12 | 292 |
1168 |
700 – 800 |
3 | 750 | 4 | 12 | 392 | 1176 |
∑ fᵢ = 50 |
7896 |
Mean x̄ \(={{A}_{i}}+\frac{\sum{{{f}_{i}}{{d}_{i}}}}{\sum{{{f}_{i}}}}\times h\),
\(=350+\frac{4}{50}\times 100=350+8\),
x̄ = 358,
Mean deviation about the mean \(=\frac{\sum{{{f}_{i}}|{{x}_{i}}-\bar{x}|}}{\sum{{{f}_{i}}}}=\frac{7896}{50}=157.92\).