A coin is tossed 20 times and X is the number of heads which occur.
what is the mean and variance of the X-distribution.
Please explain how you got the solution and don't answer if you're just
guessing.
#1
mean: 10, variance: 6
#2
mean: 10, variance: 5
#3
mean: 10, variance: 4
#4
mean: 15, variance: 6
Answers
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Let a random variable X have a binomial distribution with mean 8 and variance 4. If P(X ≤ 2) = k/(2)¹⁶, then k is equal to :
(A) 17 (B) 137 (C) 1 (D) 121
https://brainly.in/question/16076454
Answer:
\textbf{Given:}Given:
\text{Number of trials, n=20}Number of trials, n=20
\text{p=P(getting head)=$\frac{1}{2}$}p=P(getting head)=21
\text{q=P(not getting head)=$\frac{1}{2}$}q=P(not getting head)=21
\text{Since the n is finite and p is constant, the given distribution is}Since the n is finite and p is constant, the given distribution is
\text{Binomial distribution}Binomial distribution
\textbf{Mean=np}Mean=np
\text{Mean=$20{\times}\frac{1}{2}$}Mean=20×21
\implies\textbf{Mean=10}⟹Mean=10
\textbf{Variance=npq}Variance=npq
\text{Variance=$20{\times}\frac{1}{2}{\times}\frac{1}{2}$}Variance=20×21×21
\implies\textbf{Variance=5}⟹Variance=5
\therefore\textbf{Mean=10 and Variance=5}∴Mean=10 and Variance=5
\implies\textbf{Option (2) is correct}⟹Option (2) is correct
Find more:
Let a random variable X have a binomial distribution with mean 8 and variance 4. If P(X ≤ 2) = k/(2)¹⁶, then k is equal to :
(A) 17 (B) 137 (C) 1 (D)
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