English

The Mean of the Following Frequency Data is 42, Find the Missing Frequencies X and Y If the Sum of Frequencies is 100 Find X and Y. - Mathematics

Advertisements
Advertisements

Question

The mean of the following frequency data is 42, Find the missing frequencies x and y if the sum of frequencies is 100

Class 

interval 

0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80
Frequency 7 10 x 13 y 10 14 9

Find x and y.

Solution

The given data is shown as follows:

Class interval Frequency `(f_i)` Class mark `(x_i)` `f_ix_i`
0-10 7 5 35
10-20 10 15 150
20-30 x 25 25x
30-40 13 35 455
40-50 y 45 45y
50-60 10 55 550
60-70 14 65 910
70-80 9 75 675
Total `sum f_i = 63 + x+y`   `sum f_i x_i = 2775 + 25x+45y`

Sum of the frequencies = 100

⇒ Σ𝑖 𝑓𝑖 = 100
⇒ 63 + x + y = 100
⇒ x + y = 100 – 63
⇒ x + y = 37
⇒ y = 37 – x           ……..(1)
Now, the mean of the given data is given by

x = ` (sum _(i) f_i x_i )/(sum_(i) f_i )`

⇒ 42 =`(2775+25x+45y)/100`
⇒ 4200 = 2775 + 25x + 45y
⇒ 4200 – 2775 = 25x + 45y
⇒ 1425 = 25x + 45(37 – x)          [from (1)]
⇒ 1425 = 25x + 1665 – 45x
⇒ 20x = 1665 – 1425
⇒ 20x = 240
⇒ x = 12
If x = 12, then y = 37 – 12 = 25
Thus, the value of x is 12 and y is 25.

shaalaa.com
  Is there an error in this question or solution?
Chapter 9: Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive - Exercises 1

APPEARS IN

RS Aggarwal Mathematics [English] Class 10
Chapter 9 Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive
Exercises 1 | Q 11

RELATED QUESTIONS

Find the mean of the following data, using direct method:

Class 25-35 35-45 45-55 55-65 65-75
Frequency 6 10 8 12 4

If the mean of observation \[x_1 , x_2 , . . . . , x_n is x\]  then the mean of x1 + a, x2 + a, ....., xn + a is 


The measurements (in mm) of the diameters of the head of the screws are given below :

Diameter (in mm)    no. of screws
33 - 35 9
36 - 38  21
 39 - 41 30
 42 - 44 22
 45 - 47 18

Calculate the mean diameter of the head of a screw by the ' Assumed Mean Method'.


A school has 4 sections of Chemistry in class X having 40, 35, 45 and 42 students. The mean marks obtained in Chemistry test are 50, 60, 55 and 45 respectively for the 4 sections. Determine the overall average of marks per student.


A frequency distribution of the life times of 400 T.V., picture tubes leased in tube company is given below. Find the average life of tube:

Life time (in hrs) Number of tubes
300 - 399 14
400 - 499 46
500 - 599 58
600 - 699 76
700 - 799 68
800 - 899 62
900 - 999 48
1000 - 1099 22
1100 - 1199 6

If the mean of n observation ax1, ax2, ax3,....,axn is a`bar"X"`, show that `(ax_1 - abar"X") + (ax_2 - abar"X") + ...(ax_"n" - abar"X")` = 0.


In a small scale industry, salaries of employees are given in the following distribution table:

Salary (in Rs.)

4000 - 5000

5000 - 6000

6000 - 7000

7000 - 8000

8000 - 9000

9000 - 10000

Number of employees

 20 60 100 50 80 90

Then the mean salary of the employee is?


Consider the following distribution of SO2 concentration in the air (in ppm = parts per million) in 30 localities. Find the mean SO2 concentration using assumed mean method. Also find the values of A, B and C.

Class interval Frequency (fi) Class mark (xi) di = xi - a
0.00 - 0.04 4 0.02 -0.08
0.04 - 0.08 9 0.06 A
0.08 - 0.12 9 0.10 B
0.12 - 0.16 2 0.14 0.04
0.16 - 0.20 4 0.18 C
0.20 - 0.24 2 0.22 0.12
Total `sumf_i=30`    

In calculating the mean of grouped data, grouped in classes of equal width, we may use the formula `barx = a + (sumf_i d_i)/(sumf_i)` where a is the assumed mean. a must be one of the mid-points of the classes. Is the last statement correct? Justify your answer.


Find the mean of the distribution:

Class 1 – 3 3 – 5 5 – 7 7 – 10
Frequency 9 22 27 17

Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×