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Mathematics

50 persons were examined through X-ray and observations were noted as under:

Diameter of heart (in mm)Number of patients
1205
1218
12212
1239
1246
12510

Find :

(i) The mean diameter of heart,

(ii) The median diameter of heart.

Measures of Central Tendency

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Answer

Diameter of heart (in mm) (x)Number of patients (f)fxCumulative frequency
12056005
121896813 (5 + 8)
12212146425 (13 + 12)
1239110734 (25 + 9)
124674440 (34 + 6)
12510125050 (40 + 10)
TotalΣf = 50Σfx = 6133

(i) We know that,

Mean=ΣfxΣfMean=613350Mean=122.66 mm.\Rightarrow \text{Mean} = \dfrac{\text{Σfx}}{\text{Σf}} \\[1em] \Rightarrow \text{Mean} = \dfrac{6133}{50} \\[1em] \Rightarrow \text{Mean} = 122.66 \text{ mm}.

Hence, mean diameter of heart = 122.66 mm.

(ii) Here,

Number of observations (n) = 50, which is even.

By formula,

Median=(n2)th term+(n2+1)th term2Median=(502)th term+(502+1)th term2Median=25th term+(25+1)th term2Median=25 th term+26 th term2\Rightarrow \text{Median} = \dfrac{\Big(\dfrac{\text{n}}{2}\Big) \text{th} \text{ term} + \Big(\dfrac{\text{n}}{2} + 1\Big)\text{th} \text{ term}}{2} \\[1em] \Rightarrow \text{Median} = \dfrac{\Big(\dfrac{50}{2}\Big) \text{th} \text{ term} + \Big(\dfrac{50}{2} + 1\Big)\text{th} \text{ term}}{2} \\[1em] \Rightarrow \text{Median} = \dfrac{25 \text{th} \text{ term} + (25 + 1)\text{th} \text{ term}}{2} \\[1em] \Rightarrow \text{Median} = \dfrac{25 \text{ th} \text{ term} + 26\text{ th} \text{ term}}{2}

From table,

Diameter of heart corresponding to 25th term is 122

Diameter of heart corresponding to 26th term is 123

Median=25th term+26th term2Median=122+1232Median=2452Median=122.5 mm.\Rightarrow \text{Median} = \dfrac{25 \text{th} \text{ term} + 26\text{th} \text{ term}}{2} \\[1em] \Rightarrow \text{Median} = \dfrac{122 + 123}{2} \\[1em] \Rightarrow \text{Median} = \dfrac{245}{2} \\[1em] \Rightarrow \text{Median} = 122.5 \text{ mm}.

Hence, median diameter of heart = 122.5 mm.

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