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Question 26 of 61
Q.

From the following data calculate the quartile deviation:

Marks0-1515-3030-4545-6060-7575-9090-105
No. of Students826304520174
ChseodishaCHSE Odisha Plus Two (Class 12) Commerce Board 2019Subjective· 8mImportance★★★★★est
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Q1=31.75Q_1 = 31.75, Q3=62.625Q_3 = 62.625, so quartile deviation =Q3−Q12≈15.44= \dfrac{Q_3-Q_1}{2} \approx 15.44 marks.

Cumulative frequency table (N=∑f=150N = \sum f = 150):

MarksffCumulative ff
0–1588
15–302634
30–453064
45–6045109
60–7520129
75–9017146
90–1054150

Step 1 — First quartile Q1Q_1. N4=1504=37.5\dfrac{N}{4} = \dfrac{150}{4} = 37.5. The c.f. first exceeding 37.537.5 is 6464, so Q1Q_1 lies in class 30–45 (L=30L = 30, c.f. before =34= 34, f=30f = 30, h=15h = 15):

Q1=L+N4−c.f.f×h=30+37.5−3430×15=30+3.530×15=30+1.75=31.75.Q_1 = L + \frac{\tfrac{N}{4} - c.f.}{f}\times h = 30 + \frac{37.5 - 34}{30}\times 15 = 30 + \frac{3.5}{30}\times 15 = 30 + 1.75 = 31.75.

Step 2 — Third quartile Q3Q_3. 3N4=3×1504=112.5\dfrac{3N}{4} = \dfrac{3\times150}{4} = 112.5. The c.f. first exceeding 112.5112.5 is 129129, so Q3Q_3 lies in class 60–75 (L=60L = 60, c.f. before =109= 109, f=20f = 20, h=15h = 15): …

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