Question

Difficulty: MediumCumulative Frequency and Ogive

The cumulative frequency distribution below shows the monthly electricity consumption (in kWh) recorded for 5050 households in a residential estate:

Electricity Usage (kWh)Cumulative Frequency
50\leq 505
100\leq 10015
150\leq 15035
200\leq 20045
250\leq 25050

Using linear interpolation, calculate the 60th percentile (P60P_{60}) of electricity consumption in kWh.

Answer: 137.5 kWh

Answer

137.5 kWh
The 60th percentile corresponds to the 30th observation out of 50 (0.60×50=300.60 \times 50 = 30). From the cumulative frequency table, the 30th value falls in the class interval between 100100 and 150150. Using the lower boundary of 100100, previous cumulative frequency of 1515, interval frequency of 2020, and interval width of 5050, linear interpolation gives 100+301520×50=137.5100 + \frac{30 - 15}{20} \times 50 = 137.5 kWh.

Step-by-Step Solution

1
Determine the rank for the 60th percentile
Rank position = 30th household
The 60th percentile rank corresponds to 60%60\% of the total sample size (N=50N = 50), which is 0.60×50=300.60 \times 50 = 30.
2
Identify the parameters of the percentile class interval
Lower boundary L=100L = 100, CFprev=15CF_{\text{prev}} = 15, class frequency f=20f = 20, width w=50w = 50
The cumulative frequency rises from 15 to 35 in the class interval (100,150](100, 150], so the 30th value lies within this interval.
3
Calculate the value using linear interpolation
137.5 kWh
Substitute values into P60=100+(301520)×50=100+37.5=137.5P_{60} = 100 + \left(\frac{30 - 15}{20}\right) \times 50 = 100 + 37.5 = 137.5 kWh.

Key Concept

Linear interpolation for percentiles from a cumulative frequency distribution
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