Advertisements
Advertisements
प्रश्न
The mean and standard deviation of a set of n1 observations are `barx_1` and s1, respectively while the mean and standard deviation of another set of n2 observations are `barx_2` and s2, respectively. Show that the standard deviation of the combined set of (n1 + n2) observations is given by
S.D. = `sqrt((n_1(s_1)^2 + n_2(s_2)^2)/(n_1 + n_2) + (n_1n_2 (barx_1 - barx_2)^2)/(n_1 + n_2)^2)`
उत्तर
Let xi' = 1, 2, 3, 4, ..., n1
And yj' = 1, 2, 3, 4, ..., n2
∴ `barx_1 = 1/n_1 sum_(i = 1)^n x_i`
And `barx_2 = 1/n_2 sum_(j = 1)^n y_j`
⇒ `sigma_1^2 = 1/n_1 sum_(i = 1)^(n_1) (x_i - barx_1)^2`
And `sigma_2^2 = 1/n_2 sum_(j = 1)^(n_2) (y_i - barx_2)^2`
Now mean of the combined series is given by
`barx = 1/(n_1 + n_2) [sum_(i = 1)^(n_1) + sum_(j = 1)^(n_2) y_j]`
= `(n_1 barx_1 + n_2 x_2)/(n_1 + n_2)`
Therefore, `sigma^2` of the combined series is
`sigma^2 = 1/(n_1 + n_2) [sum_(i = 1)^(n_1) (x_i - barx)^2 + sum_(j = 1)^(n_2) (y_j - barx)^2]`
Now, `sum_(i = 1)^(n_1) (x_i - barx)^2 = sum_(i = 1)^(n_1) (x_i - barx_j + bar_j - barx)^2`
= `sum_(i = 1)^(n_1) (x_i - x_j)^2 + n_1 (barx_j - barx)^2 + 2(barx_j - barx) sum_(i = 1)^(n_1) (x_i - barx_j)^2`
But `sum_(i = 1)^n (x_i - barx_i)` = 0
∵ The algebraic sum of the deviation of values of first series from their mean is zero
Also `sum_(i = 1)^(n_1) (x_i - barx)^2 = n_1s_1^2 + n_1(barx_1 - barx)^2`
= `n_1s_1^2 + n_1d_1^2`
Where `d_1 = (barx_1 - barx)`
Similarly, we have
`sum_(j = 1)^(n_2) (y_j - barx)^2 = sum_(j = 1)^(n_2) (y_j - barx_i + barx_i - barx)^2`
= `n_2s_2^2 + n_2d_2^2`
Where `d_2 = (barx_2 - barx)`
Now combined Standard Deviation (S.D.)
`sigma = sqrt((n_1(s_1^2 + d_1^2) + n_2(s_2^2 + d_2^2))/(n_1 + n_2))`
Where `d_1 = barx_1 - barx`
= `barx_1 - ((n_1barx_1 + n_2 barx_2)/(n_1 + n_2))`
= `(n_2(barx_1 - barx_2))/(n_1 + n_2)`
And `d_2 = barx_2 - barx`
= `barx_2 - ((n_1barx_1 + b_2barx_2)/(n_1 + n_2))`
= `(n_1(barx_2 - barx_1))/(n_1 + n_2)`
∴ `sigma^2 = 1/(n_1 + n_2)[n_1s_1^2 + n_2s_2^2 + (n_1n_2^2(barx_1 - barx_2)^2)/(n_1 + n_2)^2 + (n_2n_1^2(barx_2 - barx_1)^2)/(n_1 + n_2)^2]`
So, `sigma = sqrt((n_1s_1^2 + n_2s_2^2)/(n_1 + n_2) + (n_1n_2(barx_1 - barx_2)^2)/(n_1 + n_2)^2`
Hence proved.
APPEARS IN
संबंधित प्रश्न
The mean and standard deviation of marks obtained by 50 students of a class in three subjects, Mathematics, Physics and Chemistry are given below:
|
Subject |
Mathematics |
Physics |
Chemistry |
|
Mean |
42 |
32 |
40.9 |
|
Standard deviation |
12 |
15 |
20 |
Which of the three subjects shows the highest variability in marks and which shows the lowest?
The variance of 20 observations is 5. If each observation is multiplied by 2, find the variance of the resulting observations.
The mean and variance of 8 observations are 9 and 9.25 respectively. If six of the observations are 6, 7, 10, 12, 12 and 13, find the remaining two observations.
For a group of 200 candidates, the mean and standard deviations of scores were found to be 40 and 15 respectively. Later on it was discovered that the scores of 43 and 35 were misread as 34 and 53 respectively. Find the correct mean and standard deviation.
The mean and standard deviation of 20 observations are found to be 10 and 2 respectively. On rechecking it was found that an observation 8 was incorrect. Calculate the correct mean and standard deviation in each of the following cases:
(i) If wrong item is omitted
(ii) if it is replaced by 12.
Show that the two formulae for the standard deviation of ungrouped data
\[\sigma = \sqrt{\frac{1}{n} \sum \left( x_i - X \right)^2_{}}\] and
\[\sigma' = \sqrt{\frac{1}{n} \sum x_i^2 - X^2_{}}\] are equivalent, where \[X = \frac{1}{n}\sum_{} x_i\]
Find the standard deviation for the following distribution:
| x : | 4.5 | 14.5 | 24.5 | 34.5 | 44.5 | 54.5 | 64.5 |
| f : | 1 | 5 | 12 | 22 | 17 | 9 | 4 |
Find the standard deviation for the following data:
| x : | 3 | 8 | 13 | 18 | 23 |
| f : | 7 | 10 | 15 | 10 | 6 |
Calculate the mean and S.D. for the following data:
| Expenditure in Rs: | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 |
| Frequency: | 14 | 13 | 27 | 21 | 15 |
A student obtained the mean and standard deviation of 100 observations as 40 and 5.1 respectively. It was later found that one observation was wrongly copied as 50, the correct figure being 40. Find the correct mean and S.D.
Calculate the mean, median and standard deviation of the following distribution:
| Class-interval: | 31-35 | 36-40 | 41-45 | 46-50 | 51-55 | 56-60 | 61-65 | 66-70 |
| Frequency: | 2 | 3 | 8 | 12 | 16 | 5 | 2 | 3 |
The means and standard deviations of heights ans weights of 50 students of a class are as follows:
| Weights | Heights | |
| Mean | 63.2 kg | 63.2 inch |
| Standard deviation | 5.6 kg | 11.5 inch |
Which shows more variability, heights or weights?
The mean and standard deviation of marks obtained by 50 students of a class in three subjects, mathematics, physics and chemistry are given below:
| Subject | Mathematics | Physics | Chemistry |
| Mean | 42 | 32 | 40.9 |
| Standard Deviation | 12 | 15 | 20 |
Which of the three subjects shows the highest variability in marks and which shows the lowest?
The standard deviation of the data:
| x: | 1 | a | a2 | .... | an |
| f: | nC0 | nC1 | nC2 | .... | nCn |
is
Let a, b, c, d, e be the observations with mean m and standard deviation s. The standard deviation of the observations a + k, b + k, c + k, d + k, e + k is
Show that the two formulae for the standard deviation of ungrouped data.
`sigma = sqrt((x_i - barx)^2/n)` and `sigma`' = `sqrt((x^2_i)/n - barx^2)` are equivalent.
Life of bulbs produced by two factories A and B are given below:
| Length of life (in hours) |
Factory A (Number of bulbs) |
Factory B (Number of bulbs) |
| 550 – 650 | 10 | 8 |
| 650 – 750 | 22 | 60 |
| 750 – 850 | 52 | 24 |
| 850 – 950 | 20 | 16 |
| 950 – 1050 | 16 | 12 |
| 120 | 120 |
The bulbs of which factory are more consistent from the point of view of length of life?
The mean and standard deviation of some data for the time taken to complete a test are calculated with the following results:
Number of observations = 25, mean = 18.2 seconds, standard deviation = 3.25 seconds. Further, another set of 15 observations x1, x2, ..., x15, also in seconds, is now available and we have `sum_(i = 1)^15 x_i` = 279 and `sum_(i = 1)^15 x^2` = 5524. Calculate the standard derivation based on all 40 observations.
Two sets each of 20 observations, have the same standard derivation 5. The first set has a mean 17 and the second a mean 22. Determine the standard deviation of the set obtained by combining the given two sets.
If for distribution `sum(x - 5)` = 3, `sum(x - 5)^2` = 43 and total number of items is 18. Find the mean and standard deviation.
Let a, b, c, d, e be the observations with mean m and standard deviation s. The standard deviation of the observations a + k, b + k, c + k, d + k, e + k is ______.
Let x1, x2, x3, x4, x5 be the observations with mean m and standard deviation s. The standard deviation of the observations kx1, kx2, kx3, kx4, kx5 is ______.
The standard deviation of a data is ______ of any change in orgin, but is ______ on the change of scale.
