The cumulative frequency distribution of the operational lifespans (in hours) for a batch … — JAMB UTME

The cumulative frequency distribution of the operational lifespans (in hours) for a batch of $100$ precision LED modules tested in a laboratory is summarized below:

Lifespan Interval (hours)	Class Boundaries	Cumulative Frequency
$100 - 119$	$99.5 - 119.5$	$10$
$120 - 139$	$119.5 - 139.5$	$25$
$140 - 159$	$139.5 - 159.5$	$60$
$160 - 179$	$159.5 - 179.5$	$85$
$180 - 199$	$179.5 - 199.5$	$100$

Using linear interpolation for cumulative frequency distributions, calculate the 75th percentile ( $P_{75}$ ) of the lifespan of these modules in hours.

Answer: 171.5 hours

Answer

The 75th percentile of operational lifespan is 171.5 hours.

The 75th percentile rank position is

0.75 \times 100 = 75

. The percentile falls within the class boundary

159.5 - 179.5

. Substituting lower class boundary

L = 159.5

, preceding cumulative frequency

c.f. = 60

, class frequency

f = 25

, and class width

c = 20

into

P_{75} = L + \left(\frac{75 - c.f.}{f}\right) \times c

gives

159.5 + \left(\frac{15}{25}\right) \times 20 = 171.5

hours.

Step-by-Step Solution

Determine the rank position of the 75th percentile.

Rank position = 75th value out of 100.

The 75th percentile corresponds to 75% of the total frequency N = 100.

Identify the percentile class interval and extract relevant parameters.

Class interval is 159.5 - 179.5, with L = 159.5, c.f. = 60, f = 25, and c = 20.

The cumulative frequency increases from 60 to 85 across the boundary 159.5 to 179.5, which contains the 75th value.

Compute the percentile value using ogive linear interpolation.

P_75 = 159.5 + [(75 - 60) / 25] * 20 = 171.5 hours.

Applying the cumulative frequency interpolation formula yields the exact value.

Key Concept

Calculating Percentiles from Cumulative Frequency / Ogives

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