Let a random variable $\mathrm{X}$ have a Binomial distribution with mean 8 and variance 4 . If $P(X \leq 2)=\frac{K}{2^{16}}$, then $K$ is

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Aicharya · Accepted Answer

Let $\mathrm{X} \sim \mathrm{B}(\mathrm{n}, \mathrm{p})$ According to the given conditions, Mean $=n p=8$ and variance $=n p q=4$ $\Rightarrow \mathrm{p}=\mathrm{q}=\frac{1}{2}$ and $\mathrm{n}=16$ $P(X \leq 2)=\frac{K}{2^{16}}$ $\Rightarrow \mathrm{P}(\mathrm{X}=0)+\mathrm{P}(\mathrm{X}=1)+\mathrm{P}(\mathrm{X}=2)=\frac{\mathrm{K}}{2^{16}}$ $\begin{aligned} 	herefore \quad{ }^{16} \mathrm{C}_0\left(\frac{1}{2}ight)^0\left(\frac{1}{2}ight)^{16}+{ }^{16} \mathrm{C}_1\left(\frac{1}{2}ight)^1 & \left(\frac{1}{2}ight)^{15} \ & +{ }^{16} \mathrm{C}_2\left(\frac{1}{2}ight)^2\left(\frac{1}{2}ight)^{14}=\frac{\mathrm{K}}{2^{16}} \end{aligned}$ $\begin{array}{ll} 	herefore & \frac{1+16+120}{2^{16}}=\frac{K}{2^{16}} \ 	herefore & K=137 \end{array}$

Let a random variable $\mathrm{X}$ have a Binomial distribution with mean 8 and variance 4 . If $P(X \leq…

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