Math, asked by gilayalon, 11 months ago

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

Answered by MaheswariS
0

\textbf{Given:}

\text{Number of trials, n=20}

\text{p=P(getting head)=$\frac{1}{2}$}

\text{q=P(not getting head)=$\frac{1}{2}$}

\text{Since the n is finite and p is constant, the given distribution is}

\text{Binomial distribution}

\textbf{Mean=np}

\text{Mean=$20{\times}\frac{1}{2}$}

\implies\textbf{Mean=10}

\textbf{Variance=npq}

\text{Variance=$20{\times}\frac{1}{2}{\times}\frac{1}{2}$}

\implies\textbf{Variance=5}

\therefore\textbf{Mean=10 and Variance=5}

\implies\textbf{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) 121

https://brainly.in/question/16076454

Answered by dhanushree7552
0

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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