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Worked Examples · Example 6

Q.Let X denote the number of hours a person watches television during a randomly selected day. The probability that X can take the values xix_i, has the following form, where k is some unknown constant.
[!FORMULA] P(X=xi)={0.2,if xi=0kxi,if xi=1 or 2k(5−xi),if xi=30otherwiseP(X = x_i) = \begin{cases} 0.2, & \text{if } x_i = 0 \\ kx_i, & \text{if } x_i = 1 \text{ or } 2 \\ k(5 - x_i), & \text{if } x_i = 3 \\ 0 & \text{otherwise} \end{cases}

a) Find the value of k.
b) What is the probability that the person watches two hours of television on a selected day?
c) What is the probability that the person watches at least two hours of television on a selected day?
d) What is the probability that the person watches at most 2 hours of television on a selected day?
e) Calculate mathematical expectation
f) Find variance and standard deviation of random variable X
Puducherry CbseNCERTSubjective· 5mImportance★★★★★
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Use ∑P=1\sum P=1 to get k=0.16k=0.16, then read probabilities and compute E(X)=1.76E(X)=1.76, Var⁡(X)=1.2224\operatorname{Var}(X)=1.2224, σ≈1.11\sigma\approx1.11.

∑iP(X=xi)=1,E(X)=∑xiP(xi),Var⁡(X)=E(X2)−[E(X)]2,σ=Var⁡(X).\displaystyle\sum_i P(X=x_i)=1,\quad E(X)=\sum x_iP(x_i),\quad \operatorname{Var}(X)=E(X^2)-[E(X)]^2,\quad \sigma=\sqrt{\operatorname{Var}(X)}.

  1. The values are P(0)=0.2, P(1)=k, P(2)=2k, P(3)=k(5−3)=2kP(0)=0.2,\ P(1)=k,\ P(2)=2k,\ P(3)=k(5-3)=2k.
  2. (a) ∑P=1: 0.2+k+2k+2k=1⇒0.2+5k=1⇒5k=0.8⇒k=0.16.\sum P=1:\ 0.2+k+2k+2k=1\Rightarrow 0.2+5k=1\Rightarrow 5k=0.8\Rightarrow k=0.16.
xx0123
P(X=x)P(X=x)0.200.160.320.32
  1. (b) P(X=2)=2k=0.32.P(X=2)=2k=0.32.
  2. (c) P(X≥2)=P(2)+P(3)=0.32+0.32=0.64.P(X\ge2)=P(2)+P(3)=0.32+0.32=0.64.
  3. (d) P(X≤2)=P(0)+P(1)+P(2)=0.20+0.16+0.32=0.68.P(X\le2)=P(0)+P(1)+P(2)=0.20+0.16+0.32=0.68. …

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