- Find f(12).
- Find f(16).
- Find $P(x \ge 16)$.
- Find $P(x \le 15)$.
- Find $E(x)$.
- Find $var(x)$ and $\sigma$.
The main objective of this question is to find the binomial probability.
This question uses the concept of the binomial distribution to find the binomial probability. In binomial distribution, we have the probability of two possible outcomes which are failure or success in an experiment that is carried out repeatedly.
Expert Answer
Given that $p$ is $0.70$ and $n$ is $20$.
We have the formula for binomial probability:
\[f(k)=\left( \begin{array}{c} n \\ k \end{array} \right) \times p^k \times (1-p)^{n-k}\]
Where $k$ is the binomial probability and $ (\begin{array}{c} n \\ k \end{array} )$ is the total combinations.
a) To find $f(12)$, we will use the above-mentioned formula for binomial probability.
By putting the given values of $p$ and $n$, we get:
\[f(k)=\left( \begin{array}{c} 20\\ 12 \end{array} \right) \times 0.70^{12} \times (1-0.70)^{20-12}\]
\[f(k)=\left( \begin{array}{c} 20\\ 12 \end{array} \right) \times 0.70^{12} \times (0.3)^{20-12}\]
\[f(k)=\left( \begin{array}{c} 20\\ 12 \end{array} \right) \times 0.70^{12} \times (0.3)^{8}\]
\[=0.114397\]
b) Computing $f(16)$, we will be using the same formula of the binomial distribution.
Inserting the given values of $p$,$f$ and $n$, we get:
\[f(k)=\left( \begin{array}{c} 20\\ 16\end{array} \right) \times 0.70^12 \times (1-0.70)^{20-16}\]
\[f(k)=\left( \begin{array}{c} 20\\ 16\end{array} \right) \times 0.70^12 \times (0.3)^{20-16}\]
\[f(k)=\left( \begin{array}{c} 20\\ 16\end{array} \right) \times 0.70^12 \times (0.3)^{4}\]
\[=0.130421\]
c) To compute $P(X\ge16)$, we will be adding the probabilities.
\[=f(16) +f(17) + f(18) +f(19) + f(20)\]
\[=0.2375\]
d) For computing $P(X\le15)$, we will be using the compliment rule of probability.
\[=1-P(X \geqq 16)\]
\[=1-0.2375\]
\[=0.7625\]
e) For finding the mean of the binomial distribution, we have a formula:
\[\mu=np\]
\[=20 \times 0.20 \]
\[=14\]
f) For computing the variance, we have the formula:
\[\sigma^2=npq=np(1-p)\]
\[=20(0.70)(1-0.70)\]
\[=20(0.70)(0.3)\]
\[=4.2\]
Calculating the standard deviation, we have formula:
\[\sigma = \sqrt{npq}=\sqrt{np(1-p)}\]
\[\sigma =\sqrt{(20)(0.70)(1-0.70)}\]
\[\sigma =\sqrt{(20)(0.70)(0.3)}\]
\[\sigma=2.0494\]
Numerical Answer
With the given number of trials $n=20$ and $p=0.7$,we have:
$f(12)=0.114397$
$f(16)=0.130421$
$P(X \ge 16)=0.2375$
$P(X \le 16)=0.7625$
$E(x)=14$
$\sigma^2=4.2$
$\sigma=2.0494$
Example
In binomial experiment consider the number of trials, $n =30$ and $p=0.6$. Compute the following:
– Find $f(14)$.
– Find $f(18)$
Given that $p$ is $0.60$ and $n$ is $30$.
We have the formula for binomial probability:
\[f(k)=\left( \begin{array}{c} n \\ k \end{array} \right) \times p^k \times (1-p)^{n-k}\]
a) To find $f(14)$, we will use the above-mentioned formula for binomial probability.
By putting the given values of $p$ and $n$ results in:
\[f(k)=\left( \begin{array}{c} 30\\ 14 \end{array} \right) \times 0.60^{14} \times (1-0.60)^{30-14}\]
\[f(k)=\left( \begin{array}{c} 30\\ 14 \end{array} \right) \times 0.60^{14} \times (0.4)^{30-14}\]
\[f(k)=\left( \begin{array}{c} 30\\ 14 \end{array} \right) \times 0.60^{14} \times (0.4)^{16}\]
\[=\left( \begin{array}{c} 30\\ 14 \end{array} \right) \times 3.365 \times 10^{-10}\]
b) To find $f(18)$, we will use the above-mentioned formula for binomial probability.
By putting the given values of $p$ and $n$ results in:
\[f(k)=\left( \begin{array}{c} 30\\ 18 \end{array} \right) \times 0.60^{18} \times (1-0.60)^{30-18}\]
\[f(k)=\left( \begin{array}{c} 30\\ 18 \end{array} \right) \times 0.60^{18} \times (0.4)^{30-18}\]
\[f(k)=\left( \begin{array}{c} 30\\ 18 \end{array} \right) \times 0.60^{18} \times (0.4)^{12}\]
\[=\left( \begin{array}{c} 30\\ 18 \end{array} \right) \times 1.70389333\times 10^{-9}\]