हिंदी

The Mean of 100 Observations is 50 and Their Standard Deviation is 5. the Sum of All Squares of All the Observations Is(A) 50,000 (B) 250,000 (C) 252500 (D) 255000 - Mathematics

Advertisements
Advertisements

प्रश्न

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

विकल्प

  •  50,000 

  •  250,000  

  • 252500 

  • 255000          

MCQ

उत्तर

Let \[\bar{ x} \] and \[\sigma\]  be the mean and standard deviation of 100 observations, respectively.

\[\therefore x = 50, \sigma = 5\]  and n = 100
Mean,\[\bar{ x} \] = 50

\[\Rightarrow \frac{\sum_{} x_i}{100} = 50\]

\[ \Rightarrow \sum_{} x_i = 5000 . . . . . \left( 1 \right)\]

Now,
Standard deviation,

\[\sigma = 5\]

\[\Rightarrow \sqrt{\frac{\sum_{} x_i^2}{100} - \left( \frac{\sum_{} x_i}{100} \right)^2} = 5\]

\[ \Rightarrow \frac{\sum_{} x_i^2}{100} - \left( \frac{5000}{100} \right)^2 = 25 \left[ \text{ From } \left( 1 \right) \right]\]

\[ \Rightarrow \frac{\sum_{} x_i^2}{100} = 25 + 2500 = 2525\]

\[ \Rightarrow \sum_{} x_i^2 = 252500\]

Thus, the sum of all squares of all the observations is 252500.

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 32: Statistics - Exercise 32.9 [पृष्ठ ५१]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 11
अध्याय 32 Statistics
Exercise 32.9 | Q 20 | पृष्ठ ५१

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 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 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:

 227, 235, 255, 269, 292, 299, 312, 321, 333, 348.


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.

 

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

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

Find the mean and variance of frequency distribution given below:

xi: 1 ≤ < 3 3 ≤ < 5 5 ≤ < 7 7 ≤ < 10
fi: 6 4 5 1

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.  


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?

 

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 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 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 v is the variance and σ is the standard deviation, then

 


If the standard deviation of a variable X is σ, then the standard deviation of variable \[\frac{a X + b}{c}\] 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 standard deviation of first 10 natural numbers is


Find the standard deviation of the first n natural numbers.


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.


If for distribution `sum(x - 5)` = 3, `sum(x - 5)^2` = 43 and total number of items is 18. Find the mean and standard deviation.


The standard deviation of the data 6, 5, 9, 13, 12, 8, 10 is ______.


Let x1, x2, ... xn be n observations. Let wi = lxi + k for i = 1, 2, ...n, where l and k are constants. If the mean of xi’s is 48 and their standard deviation is 12, the mean of wi’s is 55 and standard deviation of wi’s is 15, the values of l and k should be ______.


If the variance of a data is 121, then the standard deviation of the data is ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×