Question

the mean of following frequency distribution is 47.2 find the missing frequency

40-43. 43-46. 46-49. 49-52. 52-55
31. 58. 60. x. 27
​

Answer 1

The value of x is 44.

Explanation:

Given :  The mean of following frequency distribution is 47.2

Class interval   Frequency(f)   Mid point (x)     fx

40-43                      31                41.5               1286.5

43-46                      58                44.5              2581

46-49                      60                47.5              2850

49-52                        x                 50.5              50.5x

52-55                       27                53.5              1444.5

$\sum f=176+x$                       $\sum fx=8162+50.5x$

Mean = $\dfrac{\sum fx}{\sum f}$

Put values , we get

$47.2=\dfrac{8162+50.5x}{176+x}$

$47.2(176+x)=8162+50.5x$

$8307.2+47.2x=8162+50.5x$

$50.5x-47.2x= 8307.2-8162$

$3.3x= 145.2$

$x= \dfrac{145.2}{3.3}=44$

Hence, the value of x is 44.

# Learn more :

If mode of the following series is 54,find of f. Class intervals:0-15, 15-30, 30-45, 45-60, 60-75, 75-90. Frequencies are 3,5,f, 16,12,7.

https://brainly.in/question/11748333

Answer 2

Answer:

first we find xi (upper limit+lower limit)

these are the following xi ,we get

41.5

44.5

47.5

50.5

53.5

by the formula of step deviation , we solve the problem

now,we find ui

ui , will we

-2

-1

0

1

2

now, we find fiui

fiui will be

-62

-58

0

f

54

now the total of fiui will be

sigma=f-66

total value of fi will be

sigma=f+176

now, we put the values in formula, formula is

X^bar=A+sigma+fiui/sigmafi multiply h

47.2=47.5+f-66/f+176multiply3

0.3/3=f-66/f+176

solving the equation we get

-f-176=10f-660

660-176=10f+f

484=11f

f=44

so, the missing frequency will be 44