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Mathematics

For the following frequency distribution, find :

(i) the median

(ii) lower quartile

(iii) upper quartile

VariateFrequency
253
318
3410
4015
4510
489
506
602

Measures of Central Tendency

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Answer

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

VariateFrequencyCumulative frequency
2533
31811
341021
401536
451046
48955
50661
60263

Here, n (no. of observations) = 63, which is odd.

(i) As n is odd,

Median=n+12th observation=63+12=642=32nd observation\therefore \text{Median} = \dfrac{n + 1}{2} \text{th observation} \\[1em] = \dfrac{63 + 1}{2} \\[1em] = \dfrac{64}{2} \\[1em] = 32 \text{nd observation}

∴ Median = 32nd observation = 40.

Hence, median = 40.

(ii) As n is odd,

Lower quartile(Q1)=n+14th observation=63+14=644=16th observation\therefore \text{Lower quartile} (Q_1) = \dfrac{n + 1}{4} \text{th observation} \\[1em] = \dfrac{63 + 1}{4} \\[1em] = \dfrac{64}{4} \\[1em] = 16 \text{th observation}

∴ Lower quartile (Q1) = 16th observation = 34.

Hence, lower quartile = 34.

(iii) As n is odd,

Upper quartile(Q3)=3(n+1)4th observation=3(63+1)4=3×644=48th observation\therefore \text{Upper quartile} (Q_3) = \dfrac{3(n + 1)}{4} \text{th observation} \\[1em] = \dfrac{3(63 + 1)}{4} \\[1em] = \dfrac{3 \times 64}{4} \\[1em] = 48 \text{th observation}

∴ Upper quartile (Q3) = 48th observation = 48.

Hence, upper quartile = 48.

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