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Mathematics

From the following frequency distribution, find :

(i) the median

(ii) lower quartile

(iii) upper quartile

(iv) inter quartile range

VariateFrequency
154
186
208
229
257
278
306

Measures of Central Tendency

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Answer

The given variates are already in ascending order. We construct the cumulative frequency table as under

VariateFrequencyCumulative frequency
1544
18610
20818
22927
25734
27842
30648

Here, n (no. of observations) = 48, which is even.

(i) As n is even,

Median=n2th observation+(n2+1) th observation2=482th observation+(482+1) th observation2= 24th observation + 25th observation2\therefore \text{Median} = \dfrac{\dfrac{n}{2} \text{th observation} + \big(\dfrac{n}{2} + 1\big)\text{ th observation}}{2} \\[1em] = \dfrac{\dfrac{48}{2} \text{th observation} + \big(\dfrac{48}{2} + 1\big)\text{ th observation}}{2} \\[1em] = \dfrac{\text{ 24th observation + 25th observation}}{2} \\[1em]

All observations from 19th to 27th are equal, each = 22.

Hence, median

=22+222=442=22.= \dfrac{22 + 22}{2} \\[1em] = \dfrac{44}{2} \\[1em] = 22.

Hence, median = 22.

(ii) As n is even,

Lower quartile(Q1)=n4th observation=484=12th observation\therefore \text{Lower quartile} (Q_1) = \dfrac{n}{4} \text{th observation} \\[1em] = \dfrac{48}{4} \\[1em] = 12 \text{th observation}

∴ Lower quartile (Q1) = 12th observation = 20.

Hence, lower quartile = 20.

(iii) As n is even,

Upper quartile(Q3)=3n4th observation=3×484=1444=36th observation\therefore \text{Upper quartile} (Q_3) = \dfrac{3n}{4} \text{th observation} \\[1em] = \dfrac{3 × 48}{4} \\[1em] = \dfrac{144}{4} \\[1em] = 36 \text{th observation}

∴ Upper quartile (Q3) = 36th observation = 27.

Hence, upper quartile = 27.

(iv) Inter quartile range = Q3 - Q1 = 27 - 20 = 7.

Hence, inter quartile range = 7.

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