Question

Calculate the Arithmetic Mean and the Median of the frequency distribution given below. Hence calcula
the Mode using the empirical relation between the three.
Class limits
130-134 135-139 140-144 145–149 150-154 155-159 160-164
Frenquency
5
15
28
24
17
10
1
1974. CT 19
IC WAT Den​

Answer 1

Solution :-

Class ------------Fi ------ Xi ------FiXi----- CF

129.5-134.5---- 5 ------ 132 ----660----- 5

134.5-139.5---- 15 ------137 --- 2055----20

139.5-144.5----28 ------142 --- 3976--- 48

144.5-149.5----24 ------147 ----3558----72

149.5-154.5---- 17 ------152 ----2584----89

154.5-159.5----10 ------157 -----1570--- 99

159.5-164.5---- 1 ------ 162 ----- 162---- 100

⅀Fi = 100 ⅀FiXi = 14565

so,

→ Arithmetic Mean = ⅀FiXi / ⅀Fi = 14565 / 100 = 145.65 (Ans.)

now, Here , n = 100 .

So,

→ (n/2) = 50 .

Then, Cumulative frequency greater than 50 is 72, corresponds to the class 144.5 - 149.5.

Therefore,

→ Class 144.5 - 149.5 is the median class.

Now,

• Median = l + [{(n/2) - cf} / f] * h

from data we have :-

• l = lower limit of median class = 144.5
• n = total frequency = 100
• cf = Cumulative frequency of class before median class = 48 .
• f = frequency of median class = 24 .
• h = size of class = 5 .

Putting all value we get :-

→ Median = 144.5 + [(50 - 48)/24] * 5

→ Median = 144.5 + (2/24) * 5

→ Median = 144.5 + (5/12)

→ Median = 144.917 (Ans.)

therefore, using the empirical relation we get,

→ Mode = 3Median - 2Mean

→ Mode = 3(144.917) - 2(145.65)

→ Mode = 434.751 - 291.3

→ Mode = 143.451 (Ans.)

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