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Worked Examples · Example 5

Q.A class XII has 20 students whose marks (out of 30) are 14, 17, 25, 14, 21, 17, 17, 19, 18, 26, 18, 17, 17, 26, 19, 21, 21, 25, 14 and 19 years. If random variable X denotes the marks of a selected student given that the probability of each student to be selected is equally likely.

a) Prepare the probability distribution of the random variable X.
b) Find mean, variance and standard deviation of X.
Puducherry CbseNCERTSubjective· 5mImportance★★★★★
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✓ Free question

Tabulate the 2020 marks by frequency (each student equally likely, P=f20P=\tfrac{f}{20}), then compute mean =19.25=19.25, variance =13.8875=13.8875, SD ≈3.73\approx3.73.

P(X=x)=f20,Xˉ=E(X)=∑fx20,Var⁡(X)=∑fx220−Xˉ2,σ=Var⁡(X).\displaystyle P(X=x)=\frac{f}{20},\quad \bar X=E(X)=\frac{\sum fx}{20},\quad \operatorname{Var}(X)=\frac{\sum fx^2}{20}-\bar X^{2},\quad \sigma=\sqrt{\operatorname{Var}(X)}.

  1. Frequencies of the 2020 marks 14,17,25,14,21,17,17,19,18,26,18,17,17,26,19,21,21,25,14,1914,17,25,14,21,17,17,19,18,26,18,17,17,26,19,21,21,25,14,19:
xxffP(X=x)=f/20P(X=x)=f/20fxfxfx2fx^2
1433/203/2042588
1755/205/20851445
1822/202/2036648
1933/203/20571083
2133/203/20631323
2522/202/20501250
2622/202/20521352
Total2013857689
  1. Mean: Xˉ=∑fx20=38520=19.25.\displaystyle \bar X=\frac{\sum fx}{20}=\frac{385}{20}=19.25.
  2. E(X2)=∑fx220=768920=384.45.\displaystyle E(X^2)=\frac{\sum fx^2}{20}=\frac{7689}{20}=384.45.
  3. Variance: Var⁡(X)=384.45−(19.25)2=384.45−370.5625=13.8875.\operatorname{Var}(X)=384.45-(19.25)^2=384.45-370.5625=13.8875.
  4. Standard deviation: σ=13.8875≈3.73.\sigma=\sqrt{13.8875}\approx 3.73.
✓Final answer

Mean Xˉ=19.25\bar X=19.25, Variance =13.8875=13.8875, Standard deviation σ≈3.73\sigma\approx 3.73.

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