Question

The following data gives the information on the observed life times (in hours) of 150 electrical
components. Find the mean, median and mode .Verify the empirical relation
Life
Time(inhrs)

0-20 20-40 40-60 60-80 80-100
frequency 15 10 35 50 40

Answer 1

Solution :-

Time ---- Fi ------ Xi ------ FiXi -------- CF

0 - 20 --- 15 ---- 10 ------- 150 --------- 15

20 - 40-- 10 ---- 30 -------300 -------- 25

40 - 60 --35 ----50 -------1750 -------- 60

60 - 80 --50 ----70 ------ 3500 ------- 110

80 - 100 -40 --- 90 -------3600 -------- 150

⅀Fi = 150 ----⅀ FiXi = 9300

so,

→ Mean = ⅀ FiXi / ⅀ Fi = 9300/150 = 62

______________

now,

from data, highest frequency class = 60 - 80 = mode class .

• Mode = l + [(f1 - f0) / (2f1 - f0 - f2)] * h .

from data we have :-

• lower limit of mode class = l = 60
• size of class = 20 = h .
• f1 = frequency of mode class = 50 .
• f0 = frequency of previous class = 35 .
• f2 = frequency of next class = 40 .

Putting all values we get :-

→ Mode = 60 + [(50 - 35) / (2*50 - 35 - 40)] * 20

→ Mode = 60 + [15/(100 - 75)] * 20

→ Mode = 60 + (15/25) * 20

→ Mode = 60 + 12

→ Mode = 72

________________

Here , n = 150 .

So,

→ (n/2) = 75 .

Then, Cumulative frequency greater than 75 is 110, corresponds to the class 60 - 80.

Therefore,

→ Class 60 - 80 is the median class.

Now,

• Median = l + [{(n/2) - cf} / f] * h

from data we have :-

• l = lower limit of median class = 60
• n = total frequency = 150
• cf = Cumulative frequency of class before median class = 60 .
• f = frequency of median class = 50 .
• h = size of class = 20 .

Putting all value we get :-

→ Median = 60 + [(75 - 60)/50] * 20

→ Median = 60 + (15/50) * 20

→ Median = 60 + 6

→ Median = 66

hence, the empirical relation between mean, median and mode is :-

→ Mode = 3Median - 2Mean

→ 72 = 3 * 66 - 2 * 62

→ 72 = 198 - 124

→ 72 ≠ 74 .

since the empirical relation between mean, median and mode does not satisfy . I think given data is incorrect .

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