Let a random variable X have a Binomial distribution with mean 8 and variance 4 . If $\mathrm{P}(x \leqslant…
Let a random variable X have a Binomial distribution with mean 8 and variance 4 . If $\mathrm{P}(x \leqslant 2)=\frac{\mathrm{k}}{2^{16}}$, then k is equal to
17
121
1
137
Solution
Let $\mathrm{X} \sim \mathrm{B}(\mathrm{n}, \mathrm{p})$
According to the given conditions, mean $=\mathrm{np}=8$ and variance $=\mathrm{npq}=4$
$\Rightarrow \mathrm{p}=\mathrm{q}=\frac{1}{2}$ and $\mathrm{n}=16$
$\mathrm{P}(\mathrm{X} \leq 2)=\frac{\mathrm{K}}{2^{16}}$
$\Rightarrow \mathrm{P}(\mathrm{X}=0)+\mathrm{P}(\mathrm{X}=1)+\mathrm{P}(\mathrm{X}=2)=\frac{\mathrm{K}}{2^{16}}$
$\therefore \quad{ }^{16} \mathrm{C}_0\left(\frac{1}{2}\right)^0\left(\frac{1}{2}\right)^{16}+{ }^{16} \mathrm{C}_1\left(\frac{1}{2}\right)^1\left(\frac{1}{2}\right)^{15}$
$+{ }^{16} \mathrm{C}_2\left(\frac{1}{2}\right)^2\left(\frac{1}{2}\right)^{14}=\frac{\mathrm{K}}{2^{16}}$
$\therefore \quad \frac{1+16+120}{2^{16}}=\frac{\mathrm{K}}{2^{16}}$
$\therefore \quad \mathrm{K}=137$