The length of power failure in minute are recorded in the following table. Find Q3, D2 and P40 and interpret the results.

Question

Suresh Chand · Accepted Answer

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.

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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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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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.