हिंदी

Coefficient of Variation of Two Distributions Are 60% and 70% and Their Standard Deviations Are 21 and 16 Respectively. What Are Their Arithmetic Means? - Mathematics

Advertisements
Advertisements

प्रश्न

Coefficient of variation of two distributions are 60% and 70% and their standard deviations are 21 and 16 respectively. What are their arithmetic means?

उत्तर

The coefficient of variation (CV) for the first distribution is 60.
The coefficient of variation (CV) for the second distribution is 70.

\[SD\left( \sigma_1 \right) = 21\]
\[SD\left( \sigma_2 \right) = 16\]

We know: \[CV = \frac{\sigma}{\bar{X}} \times 100\]

From the above formula, we get:

\[CV = \frac{\sigma}{\bar{X}} \times 100\]
\[\bar{{X_1}} = \frac{21}{60} \times 100 = 35\]
\[ \bar{{X_2}} = \frac{16}{70} \times 100 = 22 . 86\]
 
shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 32: Statistics - Exercise 32.7 [पृष्ठ ४८]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 11
अध्याय 32 Statistics
Exercise 32.7 | Q 3 | पृष्ठ ४८

संबंधित प्रश्न

Find the mean and variance for the first n natural numbers.


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 (in cm) and weight (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:

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.


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.


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.


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 data:

x : 3 8 13 18 23
f : 7 10 15 10 6

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

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

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

The weight of coffee in 70 jars is shown in the following table:                                                  

Weight (in grams): 200–201 201–202 202–203 203–204 204–205 205–206
Frequency: 13 27 18 10 1 1

Determine the variance and standard deviation of the above distribution.  


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 means and standard deviations of heights ans weights of 50 students of a class are as follows: 

  Weights Heights
Mean 63.2 kg 63.2 inch
Standard deviation 5.6 kg 11.5 inch

Which shows more variability, heights or weights?

 

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.

 

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


The mean of 100 observations is 50 and their standard deviation is 5. The sum of all squares of all the observations is 


Let x1x2, ..., 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 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.


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 ______.


Find the standard deviation of the first n natural numbers.


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.


Standard deviations for first 10 natural numbers is ______.


The standard deviation is ______to the mean deviation taken from the arithmetic mean.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×