Calculate the mean deviation about the median of the following observation :
3011, 2780, 3020, 2354, 3541, 4150, 5000
Given, Numbers of observations are given.
To Find: Calculate the Mean Deviation.
Formula Used: Mean Deviation =
Explanation: Here, Observations 3011, 2780, 3020, 2354, 3541, 4150, 5000 are Given.
Since, Median is the middle number of all the observation,
So, To Find the Median, Arrange the numbers in Ascending order, we get
2354, 2780, 3011, 3020, 3541, 4150, 5000
Therefore, The Median = 3020
Deviation |d| = |x-Median|
Now, The Mean Deviation is
Mean Deviation =
Hence, The Mean Deviation is 649.42
Calculate the mean deviation about the median of the following observation :
38, 70, 48, 34, 42, 55, 63, 46, 54, 44
Given, Numbers of observations are given.
To Find: Calculate the Mean Deviation.
Formula Used: Mean Deviation =
Explanation: Here, Observations 38, 70, 48, 34, 42, 55, 63, 46, 54, 44 are Given.
Since, Median is the middle number of all the observation,
So, To Find the Median, Arrange the numbers in Ascending order, we get
34, 38,42,44,46,48,54,55,63,70
Here the Number of observations are Even then Median =
Therefore, The Median = 47
Deviation |d| = |x-Median|
And, The number of observations is 10.
Now, The Mean Deviation is
Mean Deviation =
Hence, The Mean Deviation is 8.6
Calculate the mean deviation about the median of the following observation :
34, 66, 30, 38, 44, 50, 40, 60, 42, 51
Given, Numbers of observations are given.
To Find: Calculate the Mean Deviation.
Formula Used: Mean Deviation =
Explanation: Here, Observations 34, 66, 30, 38, 44, 50, 40, 60, 42, 51 are Given.
Since, Median is the middle number of all the observation,
So, To Find the Median, Arrange the numbers in Ascending order, we get
30, 34, 38, 40, 42, 44, 50,51, 60, 66
Here the Number of observations are Even then the middle terms are 42 and 44
Therefore, The Median = = 43
Deviation |d| = |x-Median|
And, The number of observations is 10.
Now, The Mean Deviation is
Mean Deviation =
Hence, The Mean Deviation is 8.7
Calculate the mean deviation about the median of the following observation :
22, 24, 30, 27, 29, 31, 25, 28, 41, 42
Given, Numbers of observations are given.
To Find: Calculate the Mean Deviation.
Formula Used: Mean Deviation =
Explanation: Here, Observations 22, 24, 30, 27, 29, 31, 25, 28, 41, 42are Given.
Since, Median is the middle number of all the observation,
So, To Find the Median, Arrange the numbers in Ascending order, we get
22, 24, 25, 27, 28, 29, 30, 31, 41, 42
Here the Number of observations are Even then the middle terms are 28 and 29
Therefore, The Median = = 28.5
Deviation |d| = |x-Median|
And, The number of observations is 10.
Now, The Mean Deviation is
Mean Deviation =
Hence, The Mean Deviation is 8.7
Calculate the mean deviation about the median of the following observation :
38, 70, 48, 34, 63, 42, 55, 44, 53, 47
Given, Numbers of observations are given.
To Find: Calculate the Mean Deviation.
Formula Used: Mean Deviation =
Explanation: Here, Observations 38, 70, 48, 34, 63, 42, 55, 44, 53, 47 are Given.
Since, Median is the middle number of all the observation,
So, To Find the Median, Arrange the numbers in Ascending order, we get
34, 38, 43, 44, 47, 48, 53, 55, 63, 70
Here the Number of observations are Even then the middle terms are 47 and 48.
Therefore, The Median = = 47.5
Deviation |d| = |x-Median|
And, The number of observations are 10.
Now, The Mean Deviation is
Mean Deviation =
Hence, The Mean Deviation is 8.4
Calculate the mean deviation from the mean for the following data :
4, 7, 8, 9, 10, 12, 13, 17
Given, Numbers of observations are given.
To Find: Calculate the Mean Deviation from mean.
Formula Used:
Explanation: Here, Observations 4, 7, 8, 9, 10, 12, 13, 17 are Given.
Deviation |d| = |x-Mean|
Mean =
Mean of the Given Observations =
And, The number of observations is 8.
Now, The Mean Deviation is
Mean Deviation =
Mean Deviation of the given Observations =
Hence, The Mean Deviation is 3
Calculate the mean deviation from the mean for the following data :
13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17
Given, Numbers of observations are given.
To Find: Calculate the Mean Deviation from Mean.
Formula Used: Mean Deviation =
Explanation: Here, Observations 13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17 are Given.
Deviation |d| = |x-Mean|
Mean =
Mean of the Given Observations =
And, The number of observations is 12.
Now, The Mean Deviation is
Mean Deviation =
Mean Deviation of the given Observations =
Hence, The Mean Deviation is 2.33
Calculate the mean deviation from the mean for the following data :
38, 70, 48, 40, 42, 55, 63, 46, 54, 44
Given, Numbers of observations are given.
To Find: Calculate the Mean Deviation.
Formula Used: Mean Deviation =
Explanation: Here, Observations 38, 70, 48, 40, 42, 55, 63, 46, 54, 44 are Given.
Deviation |d| = |x-Mean|
Mean =
Mean of the Given Observations =
And, The number of observations is 10.
Now, The Mean Deviation is
Mean Deviation =
Mean Deviation of the given Observations =
Hence, The Mean Deviation is 8.4
Calculate the mean deviation from the mean for the following data :
36, 72, 46, 42, 60, 45, 53, 46, 51, 49
Given, Numbers of observations are given.
To Find: Calculate the Mean Deviation.
Formula Used: Mean Deviation =
Explanation: Here, Observations 36, 72, 46, 42, 60, 45, 53, 46, 51, 49 are Given.
Deviation |d| = |x-Mean|
Mean =
Mean of the Given Observations =
And, The number of observations is 10
Now, The Mean Deviation is
Mean Deviation =
Mean Deviation of the given Observations =
Hence, The Mean Deviation is 7.4
Calculate the mean deviation from the mean for the following data :
57, 64, 43, 67, 49, 59, 44, 47, 61, 59
Given, Numbers of observations are given.
To Find: Calculate the Mean Deviation.
Formula Used: Mean Deviation =
Explanation: Here, Observations 57, 64, 43, 67, 49, 59, 44, 47, 61, 59 are Given.
Deviation |d| = |x-Mean|
Mean =
Mean of the Given Observations =
And, The number of observations is 10
Now, The Mean Deviation is
Mean Deviation =
Mean Deviation of the given Observations =
Hence, The Mean Deviation is 7.4
Calculate the mean deviation of the following income groups of five and seven members from their medians:
Given, Numbers of observations are given in two groups.
To Find: Calculate the Mean Deviation from their Median.
Formula Used: Mean Deviation =
For Group 1: Since, Median is the middle number of all the observation,
So, To Find the Median, Arrange the Income of Group 1 in Ascending order, we get
4000, 4200, 4400, 4600, 4800
Therefore, The Median = 4400
Deviation |d| = |x-Median|
Now, The Mean Deviation is
Mean Deviation =
Mean Deviation Of Group 1 =
For Group 2: Since, Median is the middle number of all the observation,
So, To Find the Median, Arrange the Income of Group 2 in Ascending order, we get
3800,4000,4200,4400,4600,4800,5800
Therefore, The Median = 4400
Deviation |d| = |x-Median|
Now, The Mean Deviation is
Mean Deviation =
Mean Deviation Of Group 2 =
Hence, The Mean Deviation of Group 1 is 240 and Group 2 is 457.14
The lengths (in cm) of 10 rods in a shop are given below :
40.0, 52.3, 55.2, 72.9, 52.8, 79.0, 32.5, 15.2, 27.9, 30.2
(i) Find the mean deviation from the median
(ii) Find the mean deviation from the mean also.
(i) Given, Numbers of observations are given .
To Find: Calculate the Mean Deviation from their Median.
Formula Used: Mean Deviation =
Explanation: Since, Median is the middle number of all the observation,
So, To Find the Median, Arrange the given length of shops in Ascending order, we get
15.2, 27.9, 30.2, 32.5, 40.0, 52.3, 52.8, 55.2, 72.9, 79.0
Here the Number of observations are Even then Median =
Therefore, The Median = 46.15
Deviation |d| = |x-Median|
Now, The Mean Deviation is
Mean Deviation =
Mean Deviation From Median =
Hence, The Mean Deviation is 16.74
(ii) Here, Observations 15.2, 27.9, 30.2, 32.5, 40.0, 52.3, 52.8, 55.2, 72.9, 79.0
are Given.
Deviation |d| = |x-Mean|
Mean =
Mean of the Given Observations =
And, The number of observations is 10
Now, The Mean Deviation is
Mean Deviation =
Mean Deviation of the given Observations =
Hence, The Mean Deviation is 16.64
In question 1(iii), (iv), (v) find the number of observations lying between and , where M.D. is the mean deviation from the mean.
(iii) 34, 66, 30, 38, 44, 50, 40, 60, 42, 51
Given, Numbers of observations are given.
To Find: Calculate the Mean Deviation from Mean
Formula Used: Mean Deviation =
Explanation: Here, Observations 34, 66, 30, 38, 44, 50, 40, 60, 42, 51 are Given.
Mean =
Mean for given data
Deviation |d| = |x-Mean|
And, The number of observations is 10.
Now, The Mean Deviation is
Mean Deviation M.D=
Now,
- M.D = 45.5-9=36.5
+ M.D = 45.5+9=54.5
So, There are total 6 observation between - M.D and + M.D
(iv)Given, Numbers of observations are given.
To Find: Calculate the Mean Deviation from Mean
Formula Used: Mean Deviation =
Explanation: Numbers of observations are 22,24,30,27,29,31,25,28,41,42.
Mean =
Mean for given data
Deviation |d| = |x-Mean|
And, The number of observations is 10.
Now, The Mean Deviation is
Mean Deviation M.D=
Now,
- M.D = 29.9-4.88= 25.02
+ M.D = 29.9+4.88=34.78
So, There are total 6 observation between - M.D and + M.D
(V)Given, Numbers of observations are given.
To Find: Calculate the Mean Deviation from Mean
Formula Used: Mean Deviation =
Explanation: Numbers of observations are 38,70,48,34,63,42,55,44,53,47.
Mean =
Mean for given data
Deviation |d| = |x-Mean|
And, The number of observations is 10.
Now, The Mean Deviation is
Mean Deviation M.D=
Now,
- M.D = 49.4-8.68=40.72
+ M.D = 49.4+8.68=58.08
So, There are total 6 observation between - M.D and + M.D
Calculate the mean deviation from the median of the following frequency distribution :
Given, Numbers of observations are given.
To Find: Calculate the Mean Deviation
Formula Used: Mean Deviation =
Explanation.
Here we have to calculate the mean deviation from the median. So,
We know, Median is the Middle term,
Therefore, Median = 61
Let xi =Heights in inches
And, fi = Number of students
N=197
Mean deviation=
Hence, The mean deviation is 1.70.
The number of telephone calls received at an exchange in 245 successive on2-minute intervals is shown in the following frequency distribution :
Compute the mean deviation about the median.
Given, Numbers of observations are given.
To Find: Calculate the Mean Deviation
Formula Used: Mean Deviation =
Explanation.
Here we have to calculate the mean deviation from the median. So,
We know, Median in the even terms ,
Therefore, Median = 4
Let xi =Number of calls
And, fi = Frequency
N = 245
Mean Deviation =
Mean deviation for given data =
Hence, The mean deviation is 1.49
Calculate the mean deviation about the median of the following frequency distribution :
Given, Numbers of observations are given.
To Find: Calculate the Mean Deviation
Formula Used: Mean Deviation =
Explanation.
Here we have to calculate the mean deviation from the median. So,
Here, N =50
Then, N/2 =
SO, The median Corresponding to 25 is 13
N = 50
Mean Deviation =
Mean deviation for given data =
Hence, The mean Deviation is 2.72.
Find the mean deviation from the mean for the following data :
Given, Numbers of observations are given.
To Find: Calculate the Mean Deviation from the mean.
Formula Used: Mean Deviation =
Explanation.
Here we have to calculate the mean deviation from Mean So,
Mean =
Mean = =9
Mean Deviation =
Mean deviation for given data =
Hence, The mean Deviation is 3.3.
Find the mean deviation from the mean for the following data :
Given, Numbers of observations are given.
To Find: Calculate the Mean Deviation from the mean.
Formula Used: Mean Deviation =
Explanation.
Here we have to calculate the mean deviation from Mean So,
Mean =
Mean = =14
Mean Deviation =
Mean deviation for given data =
Hence, The mean Deviation is 6.32.
Find the mean deviation from the mean for the following data :
Given, Numbers of observations are given.
To Find: Calculate the Mean Deviation from the mean.
Formula Used: Mean Deviation =
Explanation.
Here we have to calculate the mean deviation from Mean So,
Mean =
Mean = =14
Mean Deviation =
Mean deviation for given data =
Hence, The mean Deviation is 6.32.
Find the mean deviation from the mean for the following data :
Given, Numbers of observations are given.
To Find: Calculate the Mean Deviation from the mean.
Formula Used: Mean Deviation =
Explanation.
Here we have to calculate the mean deviation from Mean So,
Mean =
Mean of given Data = =50
Mean Deviation =
Mean deviation for given data =
Hence, The mean Deviation is 16
Find the mean deviation from the mean for the following data :
Given, Numbers of observations are given.
To Find: Calculate the Mean Deviation from the mean.
Formula Used: Mean Deviation =
Explanation.
Here we have to calculate the mean deviation from Mean So,
Mean =
Mean of the given data is =21.65
Mean Deviation =
Mean deviation for given data is
Hence, The mean Deviation is 1.25.
Find the mean deviation from the median for the following data :
Given, Numbers of observations are given.
To Find: Calculate the Mean Deviation
Formula Used: Mean Deviation =
Explanation.
Here we have to calculate the mean deviation from the median. So,
N = 21
SO, The median Corresponding to 10.5 is 27
Mean Deviation =
Mean deviation for given data =
Hence, The Mean Deviation is 4.80.
Find the mean deviation from the median for the following data :
Given, Numbers of observations are given.
To Find: Calculate the Mean Deviation
Formula Used: Mean Deviation =
Explanation.
Here we have to calculate the mean deviation from the median. So,
N = 50
SO, The median Corresponding to 12.5 is 74
Mean Deviation =
Mean deviation for given data =
Hence, The Mean Deviation is 12.5.
Find the mean deviation from the median for the following data :
Given, Numbers of observations are given.
To Find: Calculate the Mean Deviation
Formula Used: Mean Deviation =
Explanation.
Here we have to calculate the mean deviation from the median. So,
N = 20
SO, The median Corresponding to 10 is 12
Mean Deviation =
Mean deviation for given data
Hence, The Mean Deviation is 1.25.
Compute the mean deviation from the median of the following distribution :
Given, Numbers of observations are given.
To Find: Calculate the Mean Deviation
Formula Used: Mean Deviation =
Explanation.
Here we have to calculate the mean deviation from the median. So,
Median is the middle term of the Xi,
Here, The middle term is 25
Therefore, Median = 25
Mean Deviation =
Mean deviation for given data
Hence, The Mean Deviation is 9
Find the mean deviation from the mean for the following data :
Given, Numbers of observations are given.
To Find: Calculate the Mean Deviation
Formula Used: Mean Deviation =
Explanation.
Here we have to calculate the mean deviation from the mean. So,
Mean =
Here, Mean =
Therefore, Mean = 358
Mean Deviation =
Mean deviation for given data
Hence, The Mean Deviation is 157.92
Find the mean deviation from the mean for the following data :
Given, Numbers of observations are given.
To Find: Calculate the Mean Deviation
Formula Used: Mean Deviation =
Explanation.
Here we have to calculate the mean deviation from the mean. So,
Mean =
Here, Mean =
Therefore, Mean = 49
Mean Deviation =
Mean deviation for given data
Find the mean deviation from the mean for the following data :
Given, Numbers of observations are given.
To Find: Calculate the Mean Deviation
Formula Used: Mean Deviation =
Explanation.
Here we have to calculate the mean deviation from the mean. So,
Mean =
Here, Mean =
Therefore, Mean = 27
Mean Deviation =
Mean deviation for given data
Hence, The Mean Deviation is 10.24
Compute mean deviation from mean of the following distribution:
Given, Numbers of observations are given.
To Find: Calculate the Mean Deviation
Formula Used: Mean Deviation =
Explanation.
Here we have to calculate the mean deviation from the mean. So,
Mean =
Here, Mean =
Therefore, Mean = 49
Mean Deviation =
Mean deviation for given data
Hence, The Mean Deviation is 14.95
The age distribution of 100 life-insurance policy holders is as follows :
Calculate the mean deviation from the median age.
Given, Numbers of observations are given.
To Find: Calculate the Mean Deviation
Formula Used: Mean Deviation =
Explanation.
Here we have to calculate the mean deviation from the median. So,
N = 96
SO, The cumulative frequency just greater than 48 is 59, and the corresponding value of x is 38.25
Median = 38.25
Mean Deviation =
Mean deviation for given data
Hence, The Mean Deviation is 10.21.
Find the mean deviation from the mean and from a median of the following distribution :
Given, Numbers of observations are given.
To Find: Calculate the Mean Deviation
Formula Used: Mean Deviation =
Explanation.
Here we have to calculate the mean deviation from the median. So,
N = 50
SO, The cumulative frequency just greater than 25 is 58, and the corresponding value of x is 28
Median = 28
Now, Mean =
Mean = 27
Mean deviation from Median
And, Mean deviation from Median
Hence, The Mean Deviation from the median is 9.56 and from mean is 9.44.
Calculate mean deviation about median age for the age distribution of 100 persons given below :
Given, Numbers of observations are given.
To Find: Calculate the Mean Deviation
Formula Used: Mean Deviation =
Explanation. Here, The Given class interval does not continue, So, we subtract 0.5 from a lower limit of the class and add 0.5 to the upper limit of the class,By this Class interval remain same while the function becomes continues.
Here we have to calculate the mean deviation from the median. So,
N = 100
Thus, the cumulative frequency slightly greater than 50 is 63 and lie under class 35.5-40.5.
Median =
Median = 38
Mean Deviation =
Mean deviation for given data
Hence, The Mean Deviation is 73.5
Calculate the mean deviation about mean for the following frequency distribution :
Given, Numbers of observations are given.
To Find: Calculate the Mean Deviation
Formula Used: Mean Deviation =
Explanation.
Here we have to calculate the mean deviation from the mean. So,
Mean =
Here, Mean =
Therefore, Mean = 9.2
Mean Deviation =
Mean deviation for given data
Hence, The Mean Deviation is 3.84
Calculate mean deviation from the median of the following data :
Given, Numbers of observations are given.
To Find: Calculate the Mean Deviation
Formula Used: Mean Deviation =
Explanation.
Here we have to calculate the mean deviation from the median. So,
N = 20
So, the cumulative frequency just greater than 10 is 12, and the corresponding value of x is 15
Median = 15
Now, Mean =
Mean = 14.1
Mean deviation from Median
And, Mean deviation from Median
Hence, The Mean Deviation from the median is 6.9 and from mean is 6.99.
Find the mean, variance and standard deviation for the following data :
2, 4, 5, 6, 8, 17
Explanation: Here, Mean
And, n=6
Variance (X) =
Variance = 23.33
Standard deviation =
Standard deviation = 4.83
Find the mean, variance and standard deviation for the following data :
6, 7, 10, 12, 13, 4, 8, 12
Explanation: Here, Mean
And, n=8
Variance (X) =
Variance = 9.25
Standard deviation =
Standard deviation = 3.04
Find the mean, variance and standard deviation for the following data :
227, 235, 255, 269, 292, 312, 321, 333, 348
Explanation: Here, Mean
And, n=10
Variance (X) =
Variance = 15394.9
Standard deviation =
Standard deviation = 39.24
Find the mean, variance and standard deviation for the following data :
15, 22, 27, 11, 9, 21, 14, 9
Explanation: Here, Mean
And, n=8
Variance (X) =
Variance = 38.75
Standard deviation =
Standard deviation = 6.22
The variance of 20 observations is 4. If each observation is multiplied by 2, find the variance of the resulting observations.
Given, The variance of 20 observations is 4.
To Find: Find the variance of resulting observations.
Explanation: Let Assume, x1,x2,x3,…,x20 be the given observations.
So, Variance (X) = 5 (Given)
X =
Now, Let u1,u2,…u20 be the new observation,
When we multiply the new observation by 2, then
Ui=2xi (for i=1,2,3…,20) ---(i)
Now,
Mean =
Mean =
Since, ui- = 2xi
2(xi)
Now, (ui-)2 = (2(xi))2
4(xi)2
Comparing Both the observations,
Variance (U) =4×Variance (X)
4×5
20
Hence, The variance of new observations is 20.
The variance of 15 observations is 4. If each observation is increased by 9, find the variance of the resulting observations.
Given, The variance of 20 observations is 4.
To Find: Find the variance of resulting observations.
Explanation: Let Assume, x1,x2,x3,…,x15 be the given observations.
So, Variance (X) = 4 (Given)
X =
Now, Let u1,u2,…u20 be the new observation,
When new observation increase by 9, then
Ui=xi+9(for i=1,2,3…,20) ---(i)
Now,
Variance (U) =60
Hence, The variance of new observations is 60.
The mean of 5 observations is 4.4 and their variance is 8.24. If three of the observations are 1, 2 and 6, find the other two observations.
Given, Mean of 5 observation is 4.4 and variance is 8.24.
To Find: Find the other two observation
Assumption: Let x and y be the other two observation. And Mean is 4.4
Here, Mean =
9+x+y=22
X+y=13 ……(1)
Now, Let Variance (X) be the variance of this observation which is to be 8.24
If is the mean than we get,
8.24 =
8.24 =
8.24 =
X2+y2 =97 ……(2)
(x+y)2+(x-y)2=2(x2+y2)
By Subsititute the value we get,
132+(x-y)2=2×97
(x-y)2=194-169
(x-y)2=25
x-y =±5 ……(3)
On solving equations (1) and (3) we get,
2x = 18
X = 9
And, y =4
Hence, The other two observations are 9 and 4.
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.
Given, The mean is 8 and SD is 4 for 6 observation
To Find: Find a new mean and new standard Deviation.
Formula Used: Standard Deviation ()
Explanation:
Let Assume, x1,x2,x3,…,x6 be the given observations.
So, Variance (X) = 8 (Given)
n=6
and (SD)
X =
Now, Let u1,u2,…u20 be the new observation,
When we multiply the new observation by 3, then
Ui=3xi (for i=1,2,3…,6) ……(i)
Now,
3×8 = 25
U = 24
Therefore, The Mean of new observation is 24
Now,
Standard Deviation
Since, Variance (X) =16
Variance (U) =
9×16
Hence, The mean of new observation is 24 and Standard deviation of new data is 12.
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.
Given, Mean of 8 observation is 9 and variance is 9.25.
To Find: Find the other two observation
Assumption: Let x and y be the other two observation. And Mean is 9
Here, Mean =
60+x+y=72
X+y=12 ……(1)
Now, Let Variance (X) be the variance of this observation which is to be 9.25
If is the mean than we get,
9.25 =
9.25 =
642+x2+y2=722
X2+y2 =80 ---(2)
(x+y)2+(x-y)2=2(x2+y2)
By Subsititute the value we get,
122+(x-y)2=2×80
(x-y)2=160-144
(x-y)2=14
x-y =±4 ……(3)
On solving equations (1) and (3) we get,
X= 8, 4
And y = 4,8
Hence, The other two observations are 8 and 4.
For a group of 200 candidates, the mean and the 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.
To Find: Find the correct mean and standard deviation.
Explanation: Here, n=200,
X =
Since, the score was incorrect,
Now, The sum is incorrect
Corrected
8000-7
The correct score is 7993
So, The mean of correct score =
Mean = 39.95
Now, Standard variance = 15
Since, Variance =
Variance = 255
X =
255 =
255 =
Now, the correct
-342-532+432+352
365000-1156-2809+1849+1225
Corrected Variance =
1820.54-1596.40
Corrected variance =224.14
Now, Corrected Standard Deviation =
Correct Deviation is 14.97
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?
To Find: Find the correct mean and standard deviation.
Explanation: Here, n=100,
X =
Now,
Corrected
3990
So, The mean of correct score =
Mean = 39.9
Now, Standard variance = 5.1
Since, Variance =
Variance = 26.01
X =
26.01 =
26.01 =
Corrected -50+40
Corrected
Corrected Variance =
1625.91-1592.01
Corrected variance =34 (Approx)
Now, Corrected Standard Deviation =
Correct Deviation is 5.83
Hence, The correct Mean is 39.9 and Correct SD is 5.83
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.
Given: Mean = 10
And, standard deviation = 2
We know that,
When wrong item is omitted,
Corrected mean
When it is replaced by 12,
Corrected mean
Now, for standard deviation,
Variance = (2)2 = 4
And we know that,
Now, omitting the wrong observation, we get,
Corrected Deviation
Now, replacing the observation
Corrected Deviation
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.
To Find: Find the correct mean and standard deviation.
Explanation: Here, n=100,
X =
So, Corrected
Corrected
Now, Standard variance = 3
Since, Variance =
Variance = 9
X =
9 =
9 =
Corrected
Corrected
Correct mean =
Corrected Variance =
409.22-400
Corrected variance =9.22
Now, Corrected Standard Deviation =
Correct Deviation is 3.04
Hence, The correct Mean is 20 and Correct SD is 3.04
Show that the two formula for the standard deviation of ungrouped data
are equivalent, where .
We know,
On Diving both sides by , we get
Taking square root both side
Hence, Proved
Find the standard deviation for the following distribution :
Given, The data is given in table/
To Find: Find the standard deviation
The formulaused: SD =
Explanation: Here, Mean =
Mean =
Now, N=70, ,
100[1.857-0.0987]
100[1.7583]
Standard Deviation
SD = 13.26
Hence, The standard deviation is 13.26
Table below shows the frequency f with which ‘x’ alpha particles were radiated from a diskette
Calculate the mean and variance.
Given, The data is given in table
To Find: Find the mean and variance
Explanation: Mean =
Mean =
Now, N=70
Hence, The mean is 3.88 and variance is 3.63
Find the mean, and standard deviation for the following data :
Given, The data is given in table/
To Find: Find the standard deviation
The formulaused: SD =
Explanation:
Now, N=109, ,
Mean
100×2.396
Variance = 239.6
Standard Deviation
SD = 15.47 years
Hence, The standard deviation is 15.47
Find the mean, and standard deviation for the following data :
Given, The data is given in table/
To Find: Find the standard deviation
The formulaused: SD =
Explanation:
Now, N=40, ,
Mean
Variance = 8.12
Standard Deviation
SD = 2.85 years
Hence, The standard deviation is 2.85
Find the standard deviation for the following data :
Given, The data is given in table/
To Find: Find the standard deviation
The formulaused: SD =
Explanation:
Variance = 37.44
Standard Deviation
SD = 6.12
Hence, The standard deviation is 6.12
Find the standard deviation for the following data :
Given, The data is given in table/
To Find: Find the standard deviation
The formulaused: SD =
Explanation:
Now, N=60, ,
Mean
Variance = 1.88
Standard Deviation
SD = 1.37
Hence, The standard deviation is 1.37
Calculate the mean and S.D. for the following data:
Given, The data is given in table/
To Find: Find the standard deviation
The formulaused: SD =
Explanation:
Mean
Mean = 26.11
Variance = 165.47
Standard Deviation
SD = 12.86
Hence, The standard deviation is 12.86
Calculate the standard deviation for the following data:
Given, The data is given in table/
To Find: Find the standard deviation
The formulaused: SD =
Explanation:
Now, N=300, ,
Mean
Variance = 1807.31
Standard Deviation
SD = 42.51
Hence, The standard deviation is 42.51
Calculate the A.M. and S.D. for the following distribution:
Given, The data is given in table/
To Find: Find the standard deviation
The formulaused: SD =
Explanation:
Now, N=79, ,
Mean
Variance = 305.20
Standard Deviation
SD = 17.47
Hence, The standard deviation is 17.47
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 is 40. Find the correct mean and S.D.
Given, Uncorrected mean is 40 and corrected SD is 5.1 and N = 100
To Find: Find the correct mean and SD
Explanation: Here,
Then,
The corrected sum of observation
So,
Now, Given Incorrect SD = 5.1
Corrected
Correct SD is 5
Hence, Correct mean is 39.90 and correct SD is 5
Calculate the mean, median and standard deviation of the following distribution
Given, The data is given in table/
To Find: Find the standard deviation
The formulaused: SD =
Explanation:
Now, N=50, ,
Mean
Variance = 58
Standard Deviation
SD = 7.62
Hence, The standard deviation is 7.62
Find the mean and variance of frequency distribution given below:
Given, The data is given in table
To Find: Find the mean and, the variance of the frequency
Explanation: Here, the class interval is not continues frequency distribution, So we have to convert into continues frequency distribution by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit then we get,
Now, N=16, ,
Mean
Variance = 3.74
Hence, The variance is 3.74
The weight of coffee in 70 jars is shown in the following table:
Determine the variance and standard deviation of the above distribution.
Given, The data is given in table
To Find: Find the mean and, the variance of the frequency
Explanation: Here, the class interval in not continues frequency distribution, So we have to convert into continues frequency distribution by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit then we get,
Now, N=70, ,
Mean
Variance = 0.98
Standard Deviation
SD = 0.99
Hence, The standard deviation is 0.99
Mean, and standard deviation of 100 observations was 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.
Given, Uncorrected mean is 40 and corrected SD is 10 and N = 100
To Find: Find the correct mean and SD
Explanation: Here,
Then,
The corrected sum of observation
So,
Variance =100
10000+160000
Incorrect
Correct
Correct
Correct SD,
Correct SD,
Hence, SD = 10.24
While calculating the mean and variance of 10 readings, a student wrongly used the reading of 52 for the correct reading 25. He obtained the mean and variance as 45 and 16 respectively. Find the correct mean and the variance.
Given, Uncorrected mean is 40 and corrected SD is 10 and N = 100
To Find: Find the correct mean and SD
Explanation: Here,
Then,
The corrected sum of observation
So,
Variance =16
Incorrect
Correct
Correct
Correct SD,
Correct SD,
SD = 6.62
Variance = 6.62×6.62
Hence, Variance = 43.82
Calculate the mean, variance and standard deviation of the following frequency distribution:
Given, The data is given in table
To Find: Find the standard deviation of the frequency
Explanation:
Now, N=70, ,
Mean
Variance = 100(3.57-1.96)
Variance = 161
Standard Deviation
SD = 12.7
Hence, The standard deviation is 12.7
Two plants A and B of a factory show the following results about the number of workers and the wages paid to them
In which plant A or B is there greater variability in individual wages?
Variation of the distribution of wages in plant A(=18)
So, Standard deviation of the distribution A()
Similarly, the Variation of the distribution of wages in plant B(=100)
So, Standard deviation of the distribution B()
And, Average monthly wages in both the plants is 2500,
Since The plant with a greater value of SD will have more variability in salary.
Hence, Plant B has more variability in individual wages than plant A
The means and standard deviations of heights and weights of 50 students in a class are as follows:
Which shows more variability, heights or weights?
Given, The mean and SD is given of 50 students.
To Find: which shows more variability, height and weight.
The formulaused: Coefficient of variations =
Explanation:
The coefficient of variations in weights =
The coefficient of variations in weights =
As results clearly show that Cv heights is greater than Cv in weights.
Hence, Heights will show more variability than weights
The coefficient of variation of two distribution are 60% and 70%, and their standard deviations are 21 and 16 respectively. What is their arithmetic means?
Here, the Coefficient of variation for the first distribution is 60
And, Coefficient of variation for the first distribution is 70
SD =21 and SD = 16
We know that, Coefficients variation =
So, Mean
For first distribution
Mean = 35
For the second distribution
Mean = 22.86
Hence, Means are 35 and 22.86 .
Calculate coefficient of variation from the following data :
Given, The data is given in table
To Find: Find the standard deviation of the frequency
Explanation:
Now, N=120,
Mean
Variance = 1076332.64
Standard Deviation
SD = 1037.47
Coefficients variation =
Cv=32.08
Hence, The coefficient variation is 332.08
An analysis of the weekly wages paid to workers in two firms A and B, belonging to the same industry gives the following results:
(i) Which firm A or B pays out the larger amount as weekly wages?
(ii) Which firm A or B has greater variability in individual wages?
(i) Average weekly wages =
Total weekly wages = (Avg weekly wages)×(No. of workers)
Total weekly wages of Firm A = 52.5×586 = Rs 30765
Total weekly wages of Firm B = 47.5×648 = Rs 30780
Firm B pays a larger amount as Firm A
(ii) Here SD(firm A) 10 and SD (Firm B) = 11
Coefficient variance (Firm A) =
Cv (Firm A) = 19.04
Coefficient variance (Firm B) =
Cv (Firm B) = 23.15
Hence, Cv of firm B is greater that that of firm A, Firm B has greater variability in individual wages.
The following are some particulars of the distribution of weights of boys and girls in a class:
Which of the distributions is more variable?
Here SD(Boys) is 3 and SD (girls) = 2
Coefficient variability =
Coefficient variance (Boys) =
Cv (Boys) = 5
Coefficient variance (Girls) =
Cv (Girls) = 4.4
Hence, Cv Boys is greater than Cv girls,then the distribution of weights of boys is more variable than that of girls
The mean and standard deviation of marks obtained by 50 students of a class in three subjects, mathematics, physics and chemistry are given below:
Which of the three subjects shows the highest variability in marks and which shows the lowest?
Coefficient variability =
Cv (Maths) =
Cv (Maths) = 28.57
Cv (physics) =
So, Cv (physics) = 46.87
Cv (chemistry) =
So, Cv (chemistry) = 48.89
Hence, Cv of chemistry is greatest, the variability of marks in chemistry is highest and that of Mathematics is lowest.
From the data given below state which group is more variable G1 or G2?
Given, The data is given in the table.
To find: Check which group is more variable G1 and G2
Explanation: Let’s find the coefficient of the variable for group G1
Here, n=150 a= 45
Mean
Variance = 227.84
Standard Deviation
Standard Deviation
The coefficient of variation =
So, Coefficient of variation =
Cv = 33.83
Now find the coefficient of variable for group G2
Here, n=150 a= 45
Mean
Variance = 243.84
Standard Deviation
Standard Deviation
The coefficient of variation =
So, Coefficient of variation =
Cv = 35.02
Since, G2 has a high coefficient of variance,
Hence, Group G2 is more variable.
Find the coefficient of variation for the following data :
Given, The data is given in table
To Find: Find the coefficient variation.
Explanation:
Now, N=100,
Mean
Variance = 37.34
Standard Deviation
SD = 6.11
Coefficients variation =
Cv=21.9
Hence, The coefficient variation is 21.9
From the prices of shares X and Y given below: Find out which is more stable in value:
Given, Data is given in the form of two table
To Find: Find out which one is more stable in value
Explanation: Let’s Find for Value X
Now, N=70
=48
SD (X)=
SD (X)==9.9
Coefficient of variation =
Coefficient of variation = = 20.6
Let’s Find for Value Y
Now, N=10
=105
SD (Y)=
SD (Y)==2
Coefficient of variation =
The coefficient of variation = = 1.90
Hence, Cv for Y is smaller that Cv of X, So X is more stable that y.
Life of bulbs produced by two factories A and B are given below:
The bulbs of which factory are more consistent from the point of view of the length of life?
For Factory A
Now, N=120,
Mean
Variance = 12097
Standard Deviation
SD = 109.98
Coefficients variation =
Cv=13.61
For Factory B
Now, N=120,
Mean
Variance = 11400
Standard Deviation
SD = 106.77
Coefficients variation =
Cv=14.29
Since, the coefficient of variation of factory B is greater than the coefficient of variation of factory A,
Hence, This means bulbs of factory A are more consistent from the point of view of the length of life.
Following are the marks obtained, out of 100, by two students Ravi and Hashina in 10 tests:
Who is more intelligent and who is more consistent?
Given, Marks obtained by two students in 10 tests are given in Table.
To Find: Who is more intelligent and who is more consistent?
Explanation: Marks obtained by Ravi
Mean,
SD ()=
SD ()=
SD ()=
SD ()=13.003
Coefficient of variation =
Cv =
For Hashima
Mean,
SD ()=
SD ()=
SD ()=
SD ()=25.16
Coefficient of variation =
Cv =
Since, The coefficient of variation in mark obtained by Hashima is greater than the coefficient of Variation in mark obtained by Ravi,
Hence, Hasima is more consistent and intelligent
Write the variance of first n natural numbers.
Let the numbers be 1,2,3,…,n
Sum of First n natural numbers is
Mean
Hence,
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.
Given, The sum of the squares of derivations for 10 observation, and mean is 2.5.
To Find: Find the standard derivation
Explanation: Here, N=10 and mean = 2.5
The square of each division =
Standard deviation
Hence, standard deviation is 0.5
If x1, x2, ……, xn are n values of a variable X and y1, y2, ……., yn are n values of variable Y such that yi = axi + b, i = 1,2, ……, n, then write Var(Y) in terms of Var(X).
And, yi=ax1+b
Hence, proved
If X and Y are two variates connected by the relation and , then write the expression for the standard deviation of Y.
Given,
To Find: Write the expression for the standard deviation of Y.
Explanation: We have
Mean (y) =
We can write as Mean (y) =
Mean (y) =
Var(X) =
But, Var(X)=
Then, Var(Y) =
Now, Substitute the value of yiand Y, then we get
Hence, Proved
In a series of 20 observations, 10 observations are each equal t k, and each of the remaining halves is equal to –k. If the standard deviation of the observation is 2, then write the value of k.
Given, n=20, di=xi-a
Where
K=±2
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.
We know,
Standard deviation
And, mean
Now, multiply by a in xi
New standard deviation,
Hence, Proved
If a variable X takes values 0, 1, 2, ….., n with frequencies nC0, nC1, nC2, …….nCn, then write variance X.
we know, mean x =
Hence,
For a frequency distribution mean deviation from mean is computed by
A.
B.
C.
D.
The general formula of Mean is
For mean deviation, d = (x-mean)
M.D =
For a frequency distribution standard deviation is computed by applying the formula
A.
B.
C.
D.
We know,
M.D =
Variance =
SD,
Hence,
If v is the variance and σ is the standard deviation, then
A.
B.
C. v = σ2
D. v2 = σ
If v is the variance and is the standard deviation, then
We know that the formula of standard variance is
So,
The mean deviation from the median is
A. equal to that measured from another value
B. Maximum if all observation are positive
C. greater than that measured from any other value.
D. less than that measured from any other value.
equal to that measured from the another value
If n = 10, and , then the coefficient of variation is
A. 36%
B. 41%
C. 25%
D. none of these
Given, n=10,
We know, Variance =
Variance =
Variance = (153-144)
SD = 3
Coefficient of variance =
Coefficient of variance =
Cv = 25
Hence, Cv = 15
The standard deviation of the data:
Is
A.
B.
C.
D. none of these
Let us calculate the mean first,
Mean
Standard deviation
The mean deviation of the series a, a+d, a+2d, ……, a+2n from its mean is
A.
B.
C.
D.
Given, Series is a, a+d, a+2d,…,+a+2n
Mean (X) =
Now, deviation from mean is xi-X
M.D =
M.D =
M.D =
Hence, M.D =
A batsman scores runs in 10 innings as 38, 70, 48, 34, 42, 55, 63, 46, 54 and 44. The mean deviation about mean is
A. 8.6
B. 6.4
C. 10.6
D. 7.6
Mean (X)=
Here, N= 10
Mean deviation =
Hence, MD is 0
The mean deviation of the numbers 3, 4, 5, 6, 7 from the mean is
A. 25
B. 5
C. 1.2
D. 0
The mean deviation of the numbers 3, 4, 5, 6, 7
Mean
=5
Mean deviation =
Mean deviation =
Hence, Mean deviation is 0
The sum of the squares deviations for 10 observations for 10 observations taken from their mean 50 is 250. The coefficient of variation is
A. 10%
B. 40%
C. 50%
D. none of these
Given, n=10 mean 250
SD,
SD = 5
Now, Coefficient of variance =
Cv =
Cv = 50
Hence, Coefficient of Variation is 10
Let x1, x2, ……xn be values taken by a variable X and y1, y2, …….yn be the values taken by a variable Y such that yi = axi + b, i = 1,2, …….,, n. Then,
A. Var (Y) = a2 Var (X)
B. Var (X) = a2 Var (Y)
C. Var (X) = Var (X) + b
D. none of these
we have given, yi=axi+b
Mean (Y) =
Mean (y) =
Then, Var(Y) =
And, Var(X) =
Hence, Var(Y) = a2 Var(X)
If the standard deviation of a variable X is σ, then the standard deviation of the variable is
A. a σ
B.
C.
D.
We have
Mean (X) =
We can write as: Mean (X) =
Mean (X) =
Var(X) =
Now, Substitute the value of yiand Y, then we get
Hence, Proved
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
A. -4
B. -8
C. 8
D. 4
Let take two observation 16 and 32
Now,
SD = 8
Now, If we divide each observation then SD is
SD = 4
Hence, SD will also be half.
If two variates X and Y are connected by the relation , where a, b, c are constants such that ac < o, then
A.
B.
C.
D. none of these
Given,
To Find: Write the expression for the standard deviation of Y.
Explanation: We have
Mean (y) =
We can write as: Mean (y) =
Mean (y) =
Var(X) =
Then, Var(Y) =
Now, Substitute the value of yiand Y, then we get
If for a sample of size 60, we have the following information and , then the variance is
A. 6.63
B. 16
C. 22
D. 44
Given, N=60,,
Variance = 44
Hence, variance = 44
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
A. s
B. ks
C. s + k
D.
Let a, b, c,d,e be the observation and mean is m
Let suppose new mean be m1
M1 =m+k
Now, The standard deviation
So, The standard deviation for new observation
Now, we can compare both observation
a+k-n=a+k-(m+k)
a+k–n = a+k–m- k
a+k-n=a-m
Similary
b+k-n=b-m
c+k-n=c-m
d+k-n=d-m
e+k-n=e-m
when we substitute the values, we get,
S1 = S
Hence, The Sd is S
The standard deviation of first 10 natural numbers is
A. 5.5
B. 3.87
C. 2.97
D. 2.87
First 10 natural numbers are 1,2,3,4,5,6,7,8,9,10
So, N =10
Hence, SD is 2.87
Consider the first 10 positive integers. If we multiply each numbers by -1 and then add 1 to each number, the variance of the numbers so obtained is
A. 8.25
B. 6.5
C. 3.87
D. 2.87
Consider 10 positive integer
Let Assume, 1,2,3,4,5,6,7,8,9,10
Now, If we multiply by -1 in each number we get,
-1,-2,-3,-4,-5,-6,-7,-8,-9,-10
And then we add 1 in each number
0,-1,-2,-3,-4,-5,-6,-7,-8,-9
Now,
Standard deviation
Hence, variance is 8.25
Consider the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. It is added to each number, the variance of the numbers so obtained is
A. 6.5
B. 2.87
C. 3.87
D. 8.25
Consider numbers are 1,2,3,4,5,6,7,8,9,10
If one is added to each number then, numbers will be
Let say xi = 2,3,4,5,6,7,8,9,10,11
So, N= 10
Standard deviation
Hence, the variance is 8.25
The mean of 100 observations is 50, and their standard deviations is 5. The sum of all squares of all the observations is
A. 50,000
B. 250,000
C. 252500
D. 255000
Given,
=50×100
=5000
Now,
Let x1, x2, ……xn, be n observations. Let yi = axi + b for I = 1, 2, ….., n, where a and b are constant. If the mean of is 48 and their standard deviation is 12, the mean of is 55 and standard deviation of is 15, the values of a and b are
A. a = 1.25, b = -5
B. a = -1.25, b = 5
C. a = 2.5, b = -5
D. a = 2.5, b = 5
Mean(y) = a.mean(x) + b
Therefore,
55 = a.48 + b
We can see that only first option satisfies this equation. Therefore, a is the correct answer.
The mean deviation of the data 3, 10, 10, 4, 7, 10, 5 from the mean is
A. 2
B. 2.57
C. 3
D. 3.57
Explanation: Mean
Mean Deviation =
Mean Deviation =
Hence, The MD is
The mean deviation for n observations x1, x2, ….., xn from their mean is given by
A.
B.
C.
D.
Let x1,x2,…xn be n observation
And X is the aithemetic mean then,
We know, Standard deviation
So,
Let x1, x2, ….., xn be n observations and be their arithmetic mean. The standard deviation is given by
A.
B.
C.
D.
Let x1,x2,…xn be n observation
And X is the arithmetic mean then,
We know, Standard deviation
So,
The standard deviation of the observations 6, 5, 9, 13, 12, 8, 10 is
A. 6
B.
C.
D.
The standard deviation of the observations 6, 5, 9, 13, 12, 8, 10 is
And, N=7
Standard deviation