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Probability — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsProbability4 Marks
Question
A discrete random variable X has the probability distribution given as below:
X
0.5
1
1.5
2
P(X)
k
k2
2k2
k
  1. Find the value of k.
  2. Determine the mean of the distribution.
✓

Answer

We have,

X
0.5
1
1.5
2
P(X)
k
k2
2k2
k
  1. We know that $\sum\limits_\text{i=1}^{\text{n}}\text{P}_{\text{i}}=1,$ where $\text{P}_{\text{i}}\geq0$

⇒ P1 + P2 + P3 + P4 = 1

⇒ k + k2 + 2k2 + k = 1

⇒ 3k2 + 2k - 1 = 0

⇒ 3k2 + 3k - k - 1 = 0

⇒ 3k(k + 1) - 1(k + 1) = 0

⇒ (3k - 1)(k + 1) = 0

$\Rightarrow\text{k}=\frac{1}{3}$ and k = -1

Since, $\text{K}\geq0,$ we take $\text{k}=\frac{1}{3}$

  1. Mean of the distribution $(\mu)=\text{E}(\text{X})=\sum\limits_{\text{i=1}}^{\text{n}}\text{x}_{\text{i}}\text{P}_{\text{i}}$

$=0.5(\text{k})+1(\text{k}^2)+1.5(2\text{k}^2)+2(\text{k})$

$=4\text{k}^2+2.5\text{k}$

$=4\cdot\frac{1}{9}+2.5\cdot\frac{1}{3}\Big[\because\text{k}=\frac{1}{3}\Big]$

$=\frac{4+7.5}{9}=\frac{23}{18}$

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