Advertisements
Advertisements
प्रश्न
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.
उत्तर
`overline x = (sumx_i)/n`
∴ `sum x_i = n overline x`
= 100 × 20
= 2000
New `sumx_i = 2000 - 21 - 21 - 18`
= 1940
New (corrected) mean = `1940/97`
= 20
σ = `1/n sqrt(n sum x_i^2 - (sumx_i)^2)`
∴ `sum x_i^2 = (n^2 σ^2 + (sumx_i)^2)/n`
= `((100)^2 xx 9 + (2000)^2)/100`
= 900 + 20 × 2000
= 900 + 40000
= 40900
Exact value of `sumx_i^2` = 40900 − (21)2 − (21)2 − (18)2
= 40900 − 441 − 441 − 324
= 39694
New (corrected) standard deviation = `1/97 sqrt(97 xx 39634 - (1940)^2)`
= `1/97 sqrt(3850318 - 3763600)`
= `1/97 xx sqrt86718`
= 3.036
APPEARS IN
संबंधित प्रश्न
Find the mean and variance for the first n natural numbers.
Find the mean and variance for the first 10 multiples of 3.
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 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
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:
- If wrong item is omitted.
- If it is replaced by 12.
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?
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.
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 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 |
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 |
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 |
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.
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 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?
Find the coefficient of variation for the following data:
| Size (in cms): | 10-15 | 15-20 | 20-25 | 25-30 | 30-35 | 35-40 |
| No. of items: | 2 | 8 | 20 | 35 | 20 | 15 |
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.
If the standard deviation of a variable X is σ, then the standard deviation of variable \[\frac{a X + b}{c}\] is
The standard deviation of first 10 natural numbers is
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.
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 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.
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.
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.
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 ______.
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.
