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Difficulty: MediumCumulative Frequency and Ogive

The table below shows the distribution of scores obtained by 6060 candidates in a competitive aptitude test:

Score Class IntervalFrequency (ff)
10−1910 - 1955
20−2920 - 291010
30−3930 - 391818
40−4940 - 491515
50−5950 - 591212

Using cumulative frequency estimation (or linear interpolation), what is the interquartile range of the distribution?

  1. 18.018.0Answer
  2. B
    20.020.0
  3. C
    30.030.0
  4. D
    15.015.0

Answer

The interquartile range of the distribution is 18.018.0.
The lower quartile (Q1Q_1) corresponds to the 15th15^{\text{th}} rank, which is exactly 29.529.5. The upper quartile (Q3Q_3) corresponds to the 45th45^{\text{th}} rank, which interpolates to 47.547.5. Subtracting Q1Q_1 from Q3Q_3 yields 47.5−29.5=18.047.5 - 29.5 = 18.0.

Step-by-Step Solution

1
Construct the cumulative frequency distribution table with upper class boundaries.
Class intervals, upper class boundaries (UCBUCB), frequencies (ff), and cumulative frequencies (cfcf):
- 10−1910 - 19: UCB=19.5UCB = 19.5, f=5f = 5, cf=5cf = 5
- 20−2920 - 29: UCB=29.5UCB = 29.5, f=10f = 10, cf=15cf = 15
- 30−3930 - 39: UCB=39.5UCB = 39.5, f=18f = 18, cf=33cf = 33
- 40−4940 - 49: UCB=49.5UCB = 49.5, f=15f = 15, cf=48cf = 48
- 50−5950 - 59: UCB=59.5UCB = 59.5, f=12f = 12, cf=60cf = 60
Cumulative frequencies and upper class boundaries are necessary to determine quartile ranks and values.
2
Find the lower quartile (Q1Q_1).
The position of Q1Q_1 is 14N=14(60)=15th\frac{1}{4} N = \frac{1}{4}(60) = 15^{\text{th}} candidate score. Since the cumulative frequency reaching UCB=29.5UCB = 29.5 is exactly 1515, Q1=29.5Q_1 = 29.5.
The 15th value lies precisely at the upper class boundary of the 20−2920 - 29 class interval.
3
Find the upper quartile (Q3Q_3) using linear interpolation.
The position of Q3Q_3 is 34N=34(60)=45th\frac{3}{4} N = \frac{3}{4}(60) = 45^{\text{th}} score.
This falls in the 40−4940 - 49 class interval (UCB=49.5UCB = 49.5, lower boundary L=39.5L = 39.5, f=15f = 15, previous cf=33cf = 33).
Q3=L+(3N4−cfprevf)×c=39.5+(45−3315)×10=39.5+8.0=47.5Q_3 = L + \left(\frac{\frac{3N}{4} - cf_{\text{prev}}}{f}\right) \times c = 39.5 + \left(\frac{45 - 33}{15}\right) \times 10 = 39.5 + 8.0 = 47.5.
Linear interpolation calculates the exact position within the target class interval.
4
Calculate the Interquartile Range (IQR).
IQR=Q3−Q1=47.5−29.5=18.0\text{IQR} = Q_3 - Q_1 = 47.5 - 29.5 = 18.0.
The interquartile range is defined as the difference between the upper quartile and the lower quartile.

Key Concept

Interquartile Range estimation from Ogive / Cumulative Frequency Distribution
Estimated Time:1m 30s
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Topics

Cumulative Frequency and Ogive

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