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BITSAT 2024
Exam
BITSAT
Year
2024
Subject
Mathematics
Chapter
Probability

BITSAT 2024 · Mathematics · Probability

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

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

Options

  1. A5
  2. B6
  3. C4
  4. DNone of these

Correct answer

C. 4

Step-by-step solution

 Mean =np=4 Variance =npq=3⇒q=34\begin{aligned} \text{ Mean } & =n p=4 \\ \text{ Variance } & =n p q=3 \\ \Rightarrow \quad q & =\frac{3}{4} \end{aligned} and p=1−q=1−34=14⇒n=16p=1-q=1-\frac{3}{4}=\frac{1}{4}\Rightarrow \quad n=16 Now, Mode of Binomial distribution is given by (n+1)P(n+1) P Case IIf (n+1)P=Integer⁡(I)(n+1) P=\operatorname{Integer}(I), then Mode ={I,I−1}=\{I, I-1\} Case II If (n+1)P≠Integer⁡(I+f)(n+1) P \neq \operatorname{Integer}(I+f), then Mode ={I}=\{I\}∴(n+1)P=(16+1)14=174=4.25=4+0.25⇒ Mode =4\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}$$
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