Advertisements
Advertisements
प्रश्न
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 |
Advertisements
उत्तर
| Class |
\[f_i\]
|
Midpoint
\[\left( x_i \right)\]
|
\[y_i = \frac{x_i - 105}{30}\]
|
\[{y_i}^2\]
|
\[f_i y_i\]
|
\[f_i {y_i}^2\]
|
| 0−30 | 9 | 15 | −3 | 9 | −27 | 81 |
| 30−60 | 17 | 45 | −2 | 4 | −34 | 68 |
| 60−90 | 43 | 75 | −1 | 1 | −43 | 43 |
| 90−120 | 82 | 105 | 0 | 0 | 0 | 0 |
| 120−150 | 81 | 135 | 1 | 1 | 81 | 81 |
| 150−180 | 44 | 165 | 2 | 4 | 88 | 176 |
| 180−210 | 24 | 195 | 3 | 9 | 72 | 216 |
|
\[\sum f_i = N = 300\]
|
\[\sum f_i y_i = 137\]
|
\[\sum f_i {y_i}^2 = 665\]
|
Mean,
\[\sigma^2 = \frac{h^2}{N^2}\left[ N\sum f_i {y_i}^2 - \left( \sum f_i y_i \right)^2 \right]\]
\[ = \frac{900}{90000}\left[ 300 \times 665 - 18769 \right]\]
\[ = \frac{1}{100}\left[ 199500 - 18769 \right]\]
\[ = \frac{180731}{100} = 1807 . 31\]
\[SD, \sigma = \sqrt{1807 . 31} = 42 . 51\]
APPEARS IN
संबंधित प्रश्न
Find the mean and variance for the first n natural numbers.
Find the mean and variance for the data.
| xi | 6 | 10 | 14 | 18 | 24 | 28 | 30 |
| fi | 2 | 4 | 7 | 12 | 8 | 4 | 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?
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?
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 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 variance of 15 observations is 4. If each observation is increased by 9, find the variance of the resulting observations.
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 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 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 |
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 |
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.
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?
From the data given below state which group is more variable, G1 or G2?
| Marks | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 | 70-80 |
| Group G1 | 9 | 17 | 32 | 33 | 40 | 10 | 9 |
| Group G2 | 10 | 20 | 30 | 25 | 43 | 15 | 7 |
If X and Y are two variates connected by the relation
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
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 standard deviation of the data 6, 5, 9, 13, 12, 8, 10 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 ______.
Standard deviations for first 10 natural numbers is ______.
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.
