THE MEDIA OF 230 OBESRVATION OF THE FOLLOWING FREQUENCY DISTRIBUTION IS 46. FIND A AND B.
CLASS
0-10
10-20
20-30
30-40
40-50
50-60
60-70
70-80
FREQUENCY
12
30
A
9
65
B
25
18
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Answers
Answer:
Class _______ frequency_______ CF
10-20 ___________12___________ 12
20-30 ___________30___________ 42
30-40_____________a__________ 42+a
40-50 ____________65_________ 107+a
50-60 ____________b__________107+a+b
60-70 ____________25_________132+a+b
70-80____________18___________150+a+b
Total observation are = 230 ( given in the question)
Median = 46
So, median class will be 40-50 (∴ 46 lies between 40-50)
Now,
150 +a+b = 230
a+b = 230-150
a+b = 80 ----------eq1
for second equation put all the values in median formula
median
l = 40
cf =42+a
f = 65
n/2 =230/2 = 115
h = 10
46-40 = (73-a)10/65
6×65 = 730-10a
-10a = -340
a = 34
now put the value of a in equation 1
b = 80- a
b = 80-34
b = 46
So, the value of a = 34 and b = 46
Answer:
Answer:
Class _______ frequency_______ CF
10-20 ___________12___________ 12
20-30 ___________30___________ 42
30-40_____________a__________ 42+a
40-50 ____________65_________ 107+a
50-60 ____________b__________107+a+b
60-70 ____________25_________132+a+b
70-80____________18___________150+a+b
Total observation are = 230 ( given in the question)
Median = 46
So, median class will be 40-50 (∴ 46 lies between 40-50)
Now,
150 +a+b = 230
a+b = 230-150
a+b = 80 ----------eq1
for second equation put all the values in median formula
median
= > l + ( \frac{n \: or \: 2 - cf}{f} )=>l+(
f
nor2−cf
)
l = 40
cf =42+a
f = 65
n/2 =230/2 = 115
h = 10
= > 40 + ( \frac{115 - (42 + a)}{65} )10=>40+(
65
115−(42+a)
)10
46-40 = (73-a)10/65
6×65 = 730-10a
-10a = -340
a = 34
now put the value of a in equation 1
b = 80- a
b = 80-34
b = 46
So, the value of a = 34 and b = 46