The length of power failure in minute are recorded in the following table.

Power failure time 22 23 24 25 26 27 28 Total
Frequency 2 5 7 10 4 3 2 33

Find Q3, D2 and P40 and interpret the results.

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Solution:

x f cf
22 2 2
23 5 7
24 7 14
25 10 24
26 4 28
27 3 31
28 2 33
N = 33

Solution For Q3:

Now, Q3 class  = \(3 \left ( \frac{N+1}{4} \right )^{th} item\)

= \(\left ( \frac{3 \enspace \times \enspace 34}{4} \right )^{th}\) = 25.5th

In cf column, the value just greater than 25.5

It mean 3/4 or 75% of power failure time has value less than or equal to 26.

Solution For D2:

Now, D2 class  = \(2 \left ( \frac{N+1}{10} \right )^{th} item\)

= \(\left ( \frac{2 \enspace \times \enspace 34}{10} \right )^{th}\) = 6.8th

In cf, the value just greater than 6.8 is 7. so, D2 = 23

Solution For P40:

Now, P40 class  = \(40 \left ( \frac{N+1}{100} \right )^{th} item\)

= \(\left ( \frac{40 \enspace \times \enspace 34}{100} \right )^{th}\) = 13.6th

In cf, the value just greater than 13.6 is 14. so, P40 = 24

It means 40% of power failure time has value less than or equal to 24.

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