Question:
Let a random variable X have a binomial distribution with mean 8 and variance 4. If
\(P(X\le2)=\frac{k}{2^{16}}\)
, then k is equal to :
Let a random variable X have a binomial distribution with mean 8 and variance 4. If
\(P(X\le2)=\frac{k}{2^{16}}\)
, then k is equal to :
\(P(X\le2)=\frac{k}{2^{16}}\)
, then k is equal to :
Show Hint
From \(np=8\) and \(npq=4\) get \(q=\frac12\), \(n=16\); then add the three binomial terms for X = 0, 1, 2.
Updated On: Nov 5, 2026
- 17
- 1
- 137
- 121
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The Correct Option is C
Solution and Explanation
\(np=8,\ npq=4\Rightarrow q=\frac12,\ p=\frac12,\ n=16\).
\(P(X\le2)=\left(\frac12\right)^{16}\left[\binom{16}{0}+\binom{16}{1}+\binom{16}{2}\right]=\frac{1+16+120}{2^{16}}=\frac{137}{2^{16}}\).
So \(k=137\).
Answer: (3) 137
\(P(X\le2)=\left(\frac12\right)^{16}\left[\binom{16}{0}+\binom{16}{1}+\binom{16}{2}\right]=\frac{1+16+120}{2^{16}}=\frac{137}{2^{16}}\).
So \(k=137\).
Answer: (3) 137
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