
How do you find the mean, variance and standard deviation of the binomial distribution with \[n=316\] and \[p=0.72\]?
Answer
509.4k+ views
Hint: For solving this question you should know about the binomial distribution. We have to find the mean, variance and standard deviation of the binomial distribution; we will find these with the help of formulas for every term. And all formulas are different for each term.
Complete step-by-step solution:
According to the question it is asked to find mean, variance and standard deviation of the binomial distribution with \[n=316\] and \[p=0.72\].
If we see the formula for binomial probability.
\[{{P}_{x}}=\left( \begin{matrix}
n \\
x \\
\end{matrix} \right){{p}^{x}}{{q}^{n-x}}\]
P = binomial probability
x = number of times for a specific outcome within n trials
\[\left( \begin{matrix}
n \\
x \\
\end{matrix} \right)=\] number of combinations
p = Probability of success on a single trail.
q = Probability of failure on a single trail.
n = number of trails
This is in probability theory and statistics, the binomial distribution with parameters n and p is the discrete probability distribution of a number of successes in a sequence of n – independent experiments.
So, the binomial distribution with \[n=316\] and \[p=0.72\].
So, if we calculate the mean, then:
\[\mu x=n.p=316\times 0.72=227.52\]
The variance is defined as \[\sigma _{x}^{2}\].
\[\Rightarrow \sigma _{x}^{2}=n.p\left( 1-p \right)\]
\[\begin{align}
& =316\times 0.72\times \left( 1-0.72 \right) \\
& =63.71 \\
\end{align}\]
And if we calculate the standard deviation, then,
\[\Rightarrow {{\sigma }_{x}}=\sqrt{n.p.\left( 1-p \right)}\]
\[\begin{align}
& =\sqrt{\left( 316.\left( 0.72 \right) \right)\left( 1-0.72 \right)} \\
& =\sqrt{63.7056} \\
& =7.98 \\
\end{align}\]
So, the mean, variance and standard deviation are 227.52, 63.71, 7.98 respectively.
Note: While solving the binomial distribution’s mean, variance and etc, which is asked in the question, always apply the right formula. And careful for the calculation. And if the question asks for the binomial expansion of any term then find that with that formula.
Complete step-by-step solution:
According to the question it is asked to find mean, variance and standard deviation of the binomial distribution with \[n=316\] and \[p=0.72\].
If we see the formula for binomial probability.
\[{{P}_{x}}=\left( \begin{matrix}
n \\
x \\
\end{matrix} \right){{p}^{x}}{{q}^{n-x}}\]
P = binomial probability
x = number of times for a specific outcome within n trials
\[\left( \begin{matrix}
n \\
x \\
\end{matrix} \right)=\] number of combinations
p = Probability of success on a single trail.
q = Probability of failure on a single trail.
n = number of trails
This is in probability theory and statistics, the binomial distribution with parameters n and p is the discrete probability distribution of a number of successes in a sequence of n – independent experiments.
So, the binomial distribution with \[n=316\] and \[p=0.72\].
So, if we calculate the mean, then:
\[\mu x=n.p=316\times 0.72=227.52\]
The variance is defined as \[\sigma _{x}^{2}\].
\[\Rightarrow \sigma _{x}^{2}=n.p\left( 1-p \right)\]
\[\begin{align}
& =316\times 0.72\times \left( 1-0.72 \right) \\
& =63.71 \\
\end{align}\]
And if we calculate the standard deviation, then,
\[\Rightarrow {{\sigma }_{x}}=\sqrt{n.p.\left( 1-p \right)}\]
\[\begin{align}
& =\sqrt{\left( 316.\left( 0.72 \right) \right)\left( 1-0.72 \right)} \\
& =\sqrt{63.7056} \\
& =7.98 \\
\end{align}\]
So, the mean, variance and standard deviation are 227.52, 63.71, 7.98 respectively.
Note: While solving the binomial distribution’s mean, variance and etc, which is asked in the question, always apply the right formula. And careful for the calculation. And if the question asks for the binomial expansion of any term then find that with that formula.
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