Question

Given mean = 67•45, find the missing frequency
X: 60-62 , 63-65 : 66-68 : 69-71 : 72-74
F: 15 , 54 , ? , 81 , 24

Answer 1

Answer:

126

Explanation:

given that mean= 67.45

let suppose the missing frequency be =x

class frequency

60-62 24

63-65 54

65-68. ?

69-71. 81

72-74. 24

first we make countinous series

because data of x is no matching with each other

take diffrence between

first data upper limit and secound data lower limit

62-63 = 1 ,than 1 is divided by 2

1÷2=0.5 this 0.5 is add upper limit and minus with lower limit

like this 60-0.5= 59.5 ,62+0.5= 62 .5 than,

class. mid value f. fx

59.5-62.5. 61 15. 915

62.5-65.5. 64. 54. 3456

65.5-68.5 67. x. 67x

68.5-71.5. 70 81 5670.

71.5-74.5. 73 24. 1752

f=174+x. fx=11793+67x

mean =67.45

so, 67.45 =11,793+67x ÷174+x do cross multiply

67.45 (174+x) =11,739+67x

11736.3+ 67.45x = 11,739+67x

11,736.3 - 11739 = 67x -67.45x

-56.7÷ -0.45 = x

126 = x

Answer 2

Answer

126

Explanation:

given that mean= 67.45

let suppose the missing frequency be =x

class frequency

60-62 24

63-65 54

65-68. ?

69-71. 81

72-74. 24

first we make countinous series

because data of x is no matching with each other

Explanation: