Advertisements
Advertisements
प्रश्न
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?
Advertisements
उत्तर
Here h = 100
Let A (assumed mean) = 800.
| Length of life (in hour) |
Mid values `(x_i)` |
`y_i = (x_i - A)/10` | Factory A | Factory B | ||||
| `f_i` | `f_iy_i` | `f_iy_i^2` | `f_i` | `f_iy_i` | `f_iy_i^2` | |||
| 550 – 650 | 600 | –2 | 10 | –20 | 40 | 8 | –16 | 32 |
| 650 – 750 | 700 | –1 | 22 | –22 | 22 | 60 | – 60 | 60 |
| 750 – 850 | 800 | 0 | 52 | 0 | 0 | 24 | 0 | 0 |
| 850 – 950 | 900 | 1 | 20 | 20 | 20 | 16 | 16 | 16 |
| 950 – 1050 | 1000 | 2 | 16 | 32 | 64 | 12 | 24 | 48 |
| 120 | 10 | 146 | 120 | –36 | 156 | |||
For factory A
Mean `(barx) = 800 + 10/120 xx 100` = 816.67 hours
S.D. = `100/120 sqrt(120(146) - 100)` = 109.98
Therefore, Coefficient of variation (C.V.) = `(S.D.)/barx xx 100`
= `109.98/816.67 xx 100`
= 13.47
For factory B
Mean = `800 + (-36)/120 100` = 770
S.D. = `100/120 sqrt(120(156) - (-36)^2)` = 110
Therefore, Coefficient of variation = `(S.D.)/"Mean" xx 100`
= `110/770 xx 100`
= 14.29
Since C.V. of factory B > C.V. of factory A
⇒ Factory B has more variability which means bulbs of factory A are more consistent.
APPEARS IN
संबंधित प्रश्न
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 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.
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:
2, 4, 5, 6, 8, 17.
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 15 observations is 4. If each observation is increased by 9, find the variance 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.
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.
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.
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 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 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.
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?
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 |
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 |
If X and Y are two variates connected by the relation
If v is the variance and σ is the standard deviation, then
The standard deviation of the data:
| x: | 1 | a | a2 | .... | an |
| f: | nC0 | nC1 | nC2 | .... | nCn |
is
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
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.
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)`
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.
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 ______.
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 ______.
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.
