Q43 / 100 · BITSAT 2024 (Memory Based Paper 3) · Mathematics

In a binomial distribution, the mean is 4 and variance is 3 . Then, its mode is

  • A
    5
  • B
    6
  • Correct answer
    C
    4
  • D
    None of these

Solution

$\begin{aligned} \text { } \text { Mean } & =n p=4 \\ \text { Variance } & =n p q=3 \\ \Rightarrow \quad q & =\frac{3}{4} \end{aligned}$ and $p=1-q=1-\frac{3}{4}=\frac{1}{4}$ $\Rightarrow \quad n=16$ Now, Mode of Binomial distribution is given by $(n+1) P$ Case IIf $(n+1) P=\operatorname{Integer}(I)$, then Mode $=\{I, I-1\}$ Case II If $(n+1) P \neq \operatorname{Integer}(I+f)$, then Mode $=\{I\}$ $\begin{aligned} \therefore(n+1) P & =(16+1) \frac{1}{4} \\ & =\frac{17}{4}=4.25 \\ & =4+0.25 \\ \Rightarrow \text { Mode } & =4 \end{aligned}$

Hence, the correct answer is option: 'C'.

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