Advertisements
Advertisements
Question
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.
Advertisements
Solution
Given that `n = 18, sum(x - 5) = 3, sum(x - 5)^2` = 43
∴ Mean = `A + (sum(x - 5))/n`
= `5 + 3/18`
= `93/18`
= 5.166
= 5.17
And S.D. = `sqrt((sum(x - 5)^2)/N - [(sum(x - 5)^2)/N]^2`
= `sqrt(43/18 - (3/18)^2`
= `sqrt(2.39 - (0.166)^2`
= `sqrt(2.39 - 0.27)`
= 1.54
Hence, the required mean is 5.17 and S.D. = 1.54
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 data.
| xi | 6 | 10 | 14 | 18 | 24 | 28 | 30 |
| fi | 2 | 4 | 7 | 12 | 8 | 4 | 3 |
The following is the record of goals scored by team A in a football session:
|
No. of goals scored |
0 |
1 |
2 |
3 |
4 |
|
No. of matches |
1 |
9 |
7 |
5 |
3 |
For the team B, mean number of goals scored per match was 2 with a standard deviation 1.25 goals. Find which team may be considered more consistent?
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
Find the mean, variance and standard deviation for the data:
6, 7, 10, 12, 13, 4, 8, 12.
Find the mean, variance and standard deviation for the data:
227, 235, 255, 269, 292, 299, 312, 321, 333, 348.
Find the mean, variance and standard deviation for the data 15, 22, 27, 11, 9, 21, 14, 9.
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.
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 data:
| x : | 3 | 8 | 13 | 18 | 23 |
| f : | 7 | 10 | 15 | 10 | 6 |
Calculate the standard deviation for the following data:
| Class: | 0-30 | 30-60 | 60-90 | 90-120 | 120-150 | 150-180 | 180-210 |
| Frequency: | 9 | 17 | 43 | 82 | 81 | 44 | 24 |
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.
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?
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?
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
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 the standard deviation of a variable X is σ, then the standard deviation of variable \[\frac{a X + b}{c}\] 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
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.
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 ______.
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)`
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 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 ______.
