Advertisements
Advertisements
प्रश्न
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.
Advertisements
उत्तर
Given that `n = 100, barx = 40, sigma = 10`
∴ `barx = (sumx_i)/N`
⇒ 40 = `(sumx_i)/100`
⇒ `sumx_i` = 400
Corrected `sumx_i` = 4000 – 30 – 70 + 3 + 27 = 3930
And Corrected mean = `3930/100` = 39.3
Now `sigma^2 = (sumx_i^2)/n - (40)^2`
⇒ 100 = `(sumx_i^2)/100 - 1600`
⇒ `sumx_i^2 = 1700 xx 100`
⇒ `sumx_i^2` = 170000
∴ Corrected `sumx_i^2 = 17000 - (30)^2 - (70)^2 + (3)^2 + (27)^2`
= 170000 – 900 – 4900 + 9 + 729
= 164938
∴ Correct S.D. = `sqrt(164938/100 - (39.3)^2`
= `sqrt(1649.38 - 1544.49)`
= `sqrt(104.89)`
= 10.24
Hence, the required S.D. = 10.24
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 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 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.
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 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 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?
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 |
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?
Coefficient of variation of two distributions are 60% and 70% and their standard deviations are 21 and 16 respectively. What are their arithmetic means?
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?
If v is the variance and σ is the standard deviation, then
If the S.D. of a set of observations is 8 and if each observation is divided by −2, the S.D. of the new set of observations will be
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
The standard deviation of the observations 6, 5, 9, 13, 12, 8, 10 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 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.
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 ______.
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 ______.
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.
The standard deviation is ______to the mean deviation taken from the arithmetic mean.
