The median of the following data is 43 1/3. Find the missing frequencies.
Class interval 0-10, 10-20, 20-30, 30-40, 40-50, 50-60, 60-70, 70-80,
frequency 5, 8, X, 12, y, 10, 14, 6 respectively. 80 = total frequency.
If you'll give the correct answer step by step, I'll mark you as a brainliast.
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Answer:
Class interval Frequency C.F.
10 - 20 12 12
20 - 30 30 42
30 - 40 f
1
42+f
1
40 - 50 65 107+f
1
50 - 60 f
2
107+f
1
+f
2
60 - 70 25 132+f
1
+f
2
70 - 80 18 150+f
1
+f
2
Let the frequency of the class 30 - 40 be f
1
and that of the calss 50 - 60 be f
2
The total frequency is 229
12 + 30 + f
1
+ 65 + f
2
+ 25 + 18 = 229 ⇒f
1
+f
2
=79
It is given that median is 46 clearly 46 lies in the class 40 - 50. So 40 - 50 is the median class
∴l=40,h=10,f=65andC=42+f
1
,N=229
Median = l+
f
2
N
−C
×h⇒46=40+
65
2
229
−(42+f
1
)
×10⇒46=40+
13
145−2f
1
⇒6=
13
145−2f
1
→2f
1
=67⇒f
1
=33.5or34 Since f
1
+f
2
=79∴f
2
=45 Hence f
1
=34 and f
2
=45
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