Exams Joint Entrance Examination Mathematics Statistics Question 16 Question 16 of 208 Medium MCQ
A random variable X takes values 0, 1, 2, 3 with probabilities 2 a + 1 30 \frac{2a+1}{30} 30 2 a + 1 , 8 a − 1 30 \frac{8a-1}{30} 30 8 a − 1 , 4 a + 1 30 \frac{4a+1}{30} 30 4 a + 1 , b b b respectively, where a , b ∈ R a, b \in \mathbb{R} a , b ∈ R . Let μ \mu μ and σ \sigma σ respectively be the mean and standard deviation of...
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X 0 1 2 3 p ( n ) 2 a + 1 30 8 a − 1 30 4 a + 1 30 b \begin{gathered} X \quad 0 \quad 1 \quad 2 \quad 3 \\ p(n) \quad \frac{2a+1}{30} \quad \frac{8a-1}{30} \quad \frac{4a+1}{30} \quad b \end{gathered} X 0 1 2 3 p ( n ) 30 2 a + 1 30 8 a − 1 30 4 a + 1 b
μ = Σ × p ( n ) = 0 + 8 a − 1 30 + 2 × 4 a + 1 30 + 3 b = 16 a + 90 b + 1 30 σ 2 = Σ x 2 p ( n ) − μ 2 ⇒ σ 2 + μ 2 = Σ x 2 p ( n ) = 2 8 a − 1 30 + 4 ( 4 a + 1 ) 30 + 9 b = 2 8 a + 90 b − 19 = 0 … … … ( i ) Σ p ( n ) = 1 ⇒ 14 a + 1 + 30 b = 30 14 a + 30 b − 29 = 0 … … . . ( i i ) a = 2 b = 1 30 = a b = 60 \begin{gathered} \mu =\Sigma \times p(n)=0+\frac{8a-1}{30}+2\times \frac{4a+1}{30}+3b=\frac{16a+90b+1}{30} \\ {\sigma }^{2}=\Sigma {x}^{2}p(n)-{\mu }^{2}\Rightarrow {\sigma }^{2}+{\mu }^{2}=\Sigma {x}^{2}p(n)=2 \\ \frac{8a-1}{30}+\frac{4(4a+1)}{30}+9b=2 \\ 8a+90b-19=0\ldots \ldots \ldots (i) \\ \Sigma p(n)=1\Rightarrow 14a+1+30b=30 \\ 14a+30b-29=0\ldots \ldots ..(ii) \\ \\ a=2\, b=\frac{1}{30}=\frac{a}{b}=60 \end{gathered} μ = Σ × p ( n ) = 0 + 30 8 a − 1 + 2 × 30 4 a + 1 + 3 b = 30 16 a + 90 b + 1 σ 2 = Σ x 2 p ( n ) − μ 2 ⇒ σ 2 + μ 2 = Σ x 2 p ( n ) = 2 30 8 a − 1 + 30 4 ( 4 a + 1 ) + 9 b = 2 8 a + 90 b − 19 = 0 ……… ( i ) Σ p ( n ) = 1 ⇒ 14 a + 1 + 30 b = 30 14 a + 30 b − 29 = 0 …… .. ( ii ) a = 2 b = 30 1 = b a = 60 Answer the question to unlock