Advertisements
Advertisements
प्रश्न
If v is the variance and σ is the standard deviation, then
विकल्प
\[v = \frac{1}{\sigma^2}\]
\[v = \frac{1}{\sigma}\]
v = σ2
v2 = σ
Advertisements
उत्तर
v = σ2
The variance is the square of the standard deviation.
APPEARS IN
संबंधित प्रश्न
Find the mean and variance for the data.
| xi | 6 | 10 | 14 | 18 | 24 | 28 | 30 |
| fi | 2 | 4 | 7 | 12 | 8 | 4 | 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 7 observations are 8 and 16, respectively. If five of the observations are 2, 4, 10, 12 and 14. 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
Given that `barx` is the mean and σ2 is the variance of n observations x1, x2, …,xn. Prove that the mean and variance of the observations ax1, ax2, ax3, …,axn are `abarx` and a2 σ2, respectively (a ≠ 0).
Find the mean, variance and standard deviation for the data:
2, 4, 5, 6, 8, 17.
The variance of 20 observations is 5. If each observation is multiplied by 2, find the variance of the resulting 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 100 observations were calculated as 40 and 5.1 respectively by a student who took by mistake 50 instead of 40 for one observation. What are the correct mean and standard deviation?
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.
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 |
Calculate the A.M. and S.D. for the following distribution:
| Class: | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 | 70-80 |
| Frequency: | 18 | 16 | 15 | 12 | 10 | 5 | 2 | 1 |
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.
Find the mean and variance of frequency distribution given below:
| xi: | 1 ≤ x < 3 | 3 ≤ x < 5 | 5 ≤ x < 7 | 7 ≤ x < 10 |
| fi: | 6 | 4 | 5 | 1 |
Mean and standard deviation of 100 observations were found to be 40 and 10 respectively. If at the time of calculation two observations were wrongly taken as 30 and 70 in place of 3 and 27 respectively, find the correct standard deviation.
Coefficient of variation of two distributions are 60% and 70% and their standard deviations are 21 and 16 respectively. What are their arithmetic means?
If X and Y are two variates connected by the relation
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
The standard deviation of first 10 natural numbers is
The standard deviation of the observations 6, 5, 9, 13, 12, 8, 10 is
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 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 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 ______.
The standard deviation of a data is ______ of any change in orgin, but is ______ on the change of scale.
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.
