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Question 60 of 61
Q.

Calculate the quartile deviation for the following distribution :

Class intervalFrequency
0 – 157
15 – 3026
30 – 4530
45 – 6045
60 – 7520
75 – 9017
90 – 1055
ChseodishaCHSE Odisha Plus Two (Class 12) Commerce Board 2024Subjective· 8mImportance★★★★★est
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Find Q1=32.25Q_1=32.25, Q3=63.375Q_3=63.375; Q.D.=Q3−Q12=15.5625Q.D.=\dfrac{Q_3-Q_1}{2}=15.5625.

Step 1 — cumulative frequencies (N=150N=150):

Classffcf
0 – 1577
15 – 302633
30 – 453063
45 – 6045108
60 – 7520128
75 – 9017145
90 – 1055150

Step 2 — Q1Q_1: N4=1504=37.5\dfrac{N}{4}=\dfrac{150}{4}=37.5. First cf ≥37.5\ge37.5 is 6363, so Q1Q_1 class =30–45=30\text{–}45 (L=30, cf=33, f=30, h=15L=30,\ cf=33,\ f=30,\ h=15):

Q1=L+N4−cff×h=30+37.5−3330×15=30+4.530×15=30+2.25=32.25.Q_1=L+\frac{\frac{N}{4}-cf}{f}\times h=30+\frac{37.5-33}{30}\times15=30+\frac{4.5}{30}\times15=30+2.25=32.25.

Step 3 — Q3Q_3: 3N4=4504=112.5\dfrac{3N}{4}=\dfrac{450}{4}=112.5. First cf ≥112.5\ge112.5 is 128128, so Q3Q_3 class =60–75=60\text{–}75 (L=60, cf=108, f=20, h=15L=60,\ cf=108,\ f=20,\ h=15): …

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