Let a random variable X have a binomial distribution with mean 8 and variance 4. If P(x le 2) = (k/216), then k is equal to :

Question

Let a random variable X have a binomial distribution with mean 8 and variance 4. If P(x le 2) = (k/216), then k is equal to : (A) 17 (B) 1 (C) 121 (D) 137

Tardigrade Admin · Accepted Answer

np = 8 npq = 4 q = (1/2) ⇒ p = (1/2) n = 16 p(x=r) = 16Cr ((1/2))16 p(x le2) = (16C0 + 16C1 +16C2/216) = (137/216)

Q. Let a random variable $X$ have a binomial distribution with mean $8$ and variance $4$ .
If $P (x \leq 2) = \frac{k}{2 ^{16}}$ , then $k$ is equal to :

17

1

121

137

$n p = 8$
$n pq = 4$
$q = \frac{1}{2} \Rightarrow p = \frac{1}{2}$
$n = 16$
$p (x = r) =^{16} C_{r} (\frac{1}{2})^{16}$
$p (x \leq 2) = \frac{^{16} C _{0} + ^{16} C _{1} + ^{16} C _{2}}{2 ^{16}}$
$= \frac{137}{2 ^{16}}$

Q. Let a random variable X have a binomial distribution with mean 8 and variance 4. If P(x≤2)=216k​, then k is equal to :

17

1

121

137

np=8 npq=4 q=21​⇒p=21​ n=16 p(x=r)=16Cr​(21​)16 p(x≤2)=21616C0​+16C1​+16C2​​ =216137​

Q. Let a random variable $X$ have a binomial distribution with mean $8$ and variance $4$ .
If $P (x \leq 2) = \frac{k}{2 ^{16}}$ , then $k$ is equal to :

$n p = 8$
$n pq = 4$
$q = \frac{1}{2} \Rightarrow p = \frac{1}{2}$
$n = 16$
$p (x = r) =^{16} C_{r} (\frac{1}{2})^{16}$
$p (x \leq 2) = \frac{^{16} C _{0} + ^{16} C _{1} + ^{16} C _{2}}{2 ^{16}}$
$= \frac{137}{2 ^{16}}$