Advertisements
Advertisements
Question
Find the mean and standard deviation using short-cut method.
| xi | 60 | 61 | 62 | 63 | 64 | 65 | 66 | 67 | 68 |
| fi | 2 | 1 | 12 | 29 | 25 | 12 | 10 | 4 | 5 |
Solution
Let the assumed mean A = 64
And yi = xi – 64
| xi | fi | yi | fiyi | `y_i^2` | `f_iy_i^2` |
| 60 | 2 | −4 | −8 | 16 | 32 |
| 61 | 1 | −3 | −3 | 9 | 9 |
| 62 | 12 | −2 | −324 | 4 | 48 |
| 63 | 29 | −1 | −329 | 1 | 29 |
| 64 | 25 | 0 | 0 | 0 | 0 |
| 65 | 12 | 1 | 12 | 1 | 12 |
| 66 | 10 | 2 | 20 | 4 | 40 |
| 67 | 4 | 3 | 12 | 9 | 36 |
| 68 | 5 | 4 | 20 | 16 | 80 |
| 100 | - | 0 | 60 | 286 |
Mean, `overlinex = A + (sumf_iy_i)/N`
= 64 + 0
= 64
Variance, σ2 = `1/N^2 [Nsumf_iy_i - (sumf_iy_i)^2]`
= `1/(100)^2 [100 xx 286 - 0]`
= `286/100`
= 2.86
Standard deviation, σ = `sqrt2.86` = 1.69
APPEARS IN
RELATED QUESTIONS
Find the mean, variance and standard deviation using short-cut method.
| Height in cms | 70 - 75 | 75 - 80 | 80 - 85 | 85 - 90 | 90 - 95 | 95 - 100 | 100 - 105 | 105 - 110 | 110 - 115 |
| No. of children | 3 | 4 | 7 | 7 | 15 | 9 | 6 | 6 | 3 |
From the prices of shares X and Y below, find out which is more stable in value:
|
X |
35 |
54 |
52 |
53 |
56 |
58 |
52 |
50 |
51 |
49 |
|
Y |
108 |
107 |
105 |
105 |
106 |
107 |
104 |
103 |
104 |
101 |
An analysis of monthly wages paid to workers in two firms A and B, belonging to the same industry, gives the following results::
|
Firm A |
Firm B |
|
|
No. of wage earners |
586 |
648 |
|
Mean of monthly wages |
Rs 5253 |
Rs 5253 |
|
Variance of the distribution of wages |
100 |
121 |
(i) Which firm A or B pays larger amount as monthly wages?
(ii) Which firm, A or B, shows greater variability in individual wages?
The frequency distribution:
| `x` | A | 2A | 3A | 4A | 5A | 6A |
| `f` | 2 | 1 | 1 | 1 | 1 | 1 |
where A is a positive integer, has a variance of 160. Determine the value of A.
For the frequency distribution:
| `x` | 2 | 3 | 4 | 5 | 6 | 7 |
| `f` | 4 | 9 | 16 | 14 | 11 | 6 |
Find the standard deviation.
There are 60 students in a class. The following is the frequency distribution of the marks obtained by the students in a test.
| Marks | 0 | 1 | 2 | 3 | 4 | 5 |
| Frequency | x – 2 | x | x2 | (x + 1)2 | 2x | x + 1 |
where x is a positive integer. Determine the mean and standard deviation of the marks.
The weights of coffee in 70 jars is shown in the following table:
| Weight (in grams) |
Frequency |
| 200 – 201 | 13 |
| 201 – 202 | 27 |
| 202 – 203 | 18 |
| 203 – 204 | 10 |
| 204 – 205 | 1 |
| 205 – 206 | 1 |
Determine variance and standard deviation of the above distribution.
Following are the marks obtained, out of 100, by two students Ravi and Hashina in 10 tests.
| Ravi | 25 | 50 | 45 | 30 | 70 | 42 | 36 | 48 | 35 | 60 |
| Hashina | 10 | 70 | 50 | 20 | 95 | 55 | 42 | 60 | 48 | 80 |
Who is more intelligent and who is more consistent?
