Advertisements
Advertisements
Question
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.
Advertisements
Solution
We have `(x_i - barx)^2 = (x_i^2 - 2barx x_i + barx^2)`
= `x_i^2 + -2barx x_i + barx^2`
= `x_i^2 - 2barx x_i + (barx)^2` 1
= `x_i^2 - 2 barx (nbarx) + n barx^2`
= `x_i^2 - nbarx^2`
Dividing both sides by n and taking their square root
We get `sigma = sigma`'.
APPEARS IN
RELATED QUESTIONS
Find the mean and variance for the data.
6, 7, 10, 12, 13, 4, 8, 12
Find the mean and variance for the first 10 multiples of 3.
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 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?
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).
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 are omitted.
Find the mean, variance and standard deviation for the data:
2, 4, 5, 6, 8, 17.
The mean and standard deviation of 6 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 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 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 |
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 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.
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.
Two plants A and B of a factory show following results about the number of workers and the wages paid to them
| Plant A | Plant B | |
| No. of workers | 5000 | 6000 |
| Average monthly wages | Rs 2500 | Rs 2500 |
| Variance of distribution of wages | 81 | 100 |
In which plant A or B is there greater variability in individual wages?
If the sum of the squares of deviations for 10 observations taken from their mean is 2.5, then write the value of standard deviation.
If X and Y are two variates connected by the relation
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.
The standard deviation of the data:
| x: | 1 | a | a2 | .... | an |
| f: | nC0 | nC1 | nC2 | .... | nCn |
is
The standard deviation of first 10 natural numbers is
The mean of 100 observations is 50 and their standard deviation is 5. The sum of all squares of all the observations is
A set of n values x1, x2, ..., xn has standard deviation 6. The standard deviation of n values x1 + k, x2 + k, ..., xn + k will be ______.
Find the standard deviation of the first n natural numbers.
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.
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 ______.
Standard deviations for first 10 natural numbers is ______.
Coefficient of variation of two distributions are 50 and 60, and their arithmetic means are 30 and 25 respectively. Difference of their standard deviation is ______.
The standard deviation of a data is ______ of any change in orgin, but is ______ on the change of scale.
