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Worked Examples · Example 2
Q.

For the following distribution of marks scored by a class of 40 students, calculate the range and quartile deviation (Q.D.).

Class interval0–1010–2020–4040–6060–90
No. of students (f)581674
Telangana TsbieTextbookSubjectiveImportance★★★★★est
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Range =90= 90; Q1=16.25Q_1 = 16.25, Q3=42.87Q_3 = 42.87, so Q.D.=13.31Q.D. = 13.31.

Step 1 — Range. Range is the difference between the upper limit of the highest class and the lower limit of the lowest class:

R=90−0=90R = 90 - 0 = 90

Step 2 — Cumulative frequencies.

Class intervalFrequency (f)Cumulative frequency (c.f.)
0–1055
10–20813
20–401629
40–60736
60–90440
Total40

Step 3 — First quartile Q1Q_1. In a continuous series Q1Q_1 is the size of the n4th=404=10th\dfrac{n}{4}\text{th} = \dfrac{40}{4} = 10\text{th} value, which lies in the class 10–20 (c.f. reaches 13 there). With L=10L = 10, c.f.=5c.f. = 5, f=8f = 8 and i=10i = 10:

Q1=L+n4−c.f.f×i=10+10−58×10=10+6.25=16.25Q_1 = L + \dfrac{\frac{n}{4} - c.f.}{f} \times i = 10 + \dfrac{10 - 5}{8} \times 10 = 10 + 6.25 = 16.25

Step 4 — Third quartile Q3Q_3. Q3Q_3 is the size of the 3n4th=3×404=30th\dfrac{3n}{4}\text{th} = \dfrac{3 \times 40}{4} = 30\text{th} value, which lies in the class 40–60 (c.f. reaches 36 there). With L=40L = 40, c.f.=29c.f. = 29, f=7f = 7 and i=20i = 20:

Q3=L+3n4−c.f.f×i=40+30−297×20=40+2.87=42.87Q_3 = L + \dfrac{\frac{3n}{4} - c.f.}{f} \times i = 40 + \dfrac{30 - 29}{7} \times 20 = 40 + 2.87 = 42.87

Step 5 — Quartile deviation.

Q.D.=Q3−Q12=42.87−16.252=26.622=13.31Q.D. = \dfrac{Q_3 - Q_1}{2} = \dfrac{42.87 - 16.25}{2} = \dfrac{26.62}{2} = 13.31

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