Advertisements
Advertisements
प्रश्न
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.
Advertisements
उत्तर
Given that `n_1 = 20, sigma_1 = 5, barx_1 = 17`
And `n_2 = 20, sigma_2 = 5, barx_2 = 22`
Now we know for combined two series that
`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`
= `sqrt((20 xx (5)^2 + 20 xx (5)^2)/(20 + 20) + (20 xx 20(17 - 22)^2)/(20 + 20)^2`
= `sqrt(1000/40 + (400 xx 25)/1600)`
= `sqrt(25 + 25/4)`
= `sqrt(125/4)`
= `sqrt(31.25)`
= 5.59
Hence, the required S.D. = 5.59
APPEARS IN
संबंधित प्रश्न
Find the mean and variance for the data.
6, 7, 10, 12, 13, 4, 8, 12
Find the mean and variance for the data.
| xi | 92 | 93 | 97 | 98 | 102 | 104 | 109 |
| fi | 3 | 2 | 3 | 2 | 6 | 3 | 3 |
The diameters of circles (in mm) drawn in a design are given below:
| Diameters | 33 - 36 | 37 - 40 | 41 - 44 | 45 - 48 | 49 - 52 |
| No. of circles | 15 | 17 | 21 | 22 | 25 |
Calculate the standard deviation and mean diameter of the circles.
[Hint: First make the data continuous by making the classes as 32.5 - 36.5, 36.5 - 40.5, 40.5 - 44.5, 44.5 - 48.5, 48.5 - 52.5 and then proceed.]
The sum and sum of squares corresponding to length x (in cm) and weight y (in gm) of 50 plant products are given below:
`sum_(i-1)^50 x_i = 212, sum_(i=1)^50 x_i^2 = 902.8, sum_(i=1)^50 y_i = 261, sum_(i = 1)^50 y_i^2 = 1457.6`
Which is more varying, the length or weight?
The mean and variance of eight 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.
The mean and standard deviation of six observations are 8 and 4, respectively. If each observation is multiplied by 3, find the new mean and new standard deviation of the resulting observations
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:
- If wrong item is omitted.
- If it is replaced by 12.
Find the mean, variance and standard deviation for the data 15, 22, 27, 11, 9, 21, 14, 9.
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.
The mean and standard deviation of a group of 100 observations were found to be 20 and 3 respectively. Later on it was found that three observations were incorrect, which were recorded as 21, 21 and 18. Find the mean and standard deviation if the incorrect observations were omitted.
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 |
The weight of coffee in 70 jars is shown in the following table:
| Weight (in grams): | 200–201 | 201–202 | 202–203 | 203–204 | 204–205 | 205–206 |
| Frequency: | 13 | 27 | 18 | 10 | 1 | 1 |
Determine the variance and standard deviation of the above distribution.
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?
Coefficient of variation of two distributions are 60% and 70% and their standard deviations are 21 and 16 respectively. What are their arithmetic means?
In a series of 20 observations, 10 observations are each equal to k and each of the remaining half is equal to − k. If the standard deviation of the observations is 2, then write the value of k.
If each observation of a raw data whose standard deviation is σ is multiplied by a, then write the S.D. of the new set of observations.
If v is the variance and σ is the standard deviation, then
The standard deviation of the data:
| x: | 1 | a | a2 | .... | an |
| f: | nC0 | nC1 | nC2 | .... | nCn |
is
If the standard deviation of a variable X is σ, then the standard deviation of variable \[\frac{a X + b}{c}\] is
The mean of 100 observations is 50 and their standard deviation is 5. The sum of all squares of all the observations is
Let x1, x2, ..., xn be n observations. Let \[y_i = a x_i + b\] for i = 1, 2, 3, ..., n, where a and b are constants. If the mean of \[x_i 's\] is 48 and their standard deviation is 12, the mean of \[y_i 's\] is 55 and standard deviation of \[y_i 's\] is 15, the values of a and b are
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?
Find the standard deviation of the first n natural numbers.
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)`
The mean life of a sample of 60 bulbs was 650 hours and the standard deviation was 8 hours. A second sample of 80 bulbs has a mean life of 660 hours and standard deviation 7 hours. Find the overall standard deviation.
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.
The standard deviation of the data 6, 5, 9, 13, 12, 8, 10 is ______.
Let x1, x2, ... xn be n observations. Let wi = lxi + k for i = 1, 2, ...n, where l and k are constants. If the mean of xi’s is 48 and their standard deviation is 12, the mean of wi’s is 55 and standard deviation of wi’s is 15, the values of l and k should be ______.
The standard deviation is ______to the mean deviation taken from the arithmetic mean.
