# Binomial distribution and finding the probability. I want to know how to use the binomial distribution table, given that the mean of a binomial distribution is 5 and the standard deviation is 2. What is the actual probability of 5 successes?

Binomial distribution and finding the probability
I want to know how to use the binomial distribution table, given that the mean of a binomial distribution is 5 and the standard deviation is 2. What is the actual probability of 5 successes? The(mean) is found by number of trials *probability of success in any trial; while the S.D. is found by square root of (number of trials) * (probability of success) * (probability of success-1). How do i use the binomial distribution table to find the probability or even without it, but without the use of permutations and combinations.
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geoforoiunpwd
Step 1
At the first step, you must calculate the value of p.
$mean=\mu =np=5\phantom{\rule{1em}{0ex}}\left(1\right)\phantom{\rule{0ex}{0ex}}sd=2⇒\sqrt{np\left(1-p\right)}=2\phantom{\rule{1em}{0ex}}\left(2\right)$
$\left(1\right),\left(2\right)⇒\frac{4}{5}=1-p⇒p=0.2\phantom{\rule{1em}{0ex}}\left(3\right)$
$\left(1\right),\left(3\right)⇒n=25$
Step 2
In the binomial distribution table, you must find a row with $n=25,x=5$ and column with $p=0.2$.
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Dymnembalmese2n
Step 1
If you don´t have a table with $n=25$ and a calculator which is able to calculate binomial coefficients directly you can decompose the binomial coefficient.
In general the asked probability is $P\left(X=k\right)=\left(\genfrac{}{}{0}{}{n}{k}\right)\cdot {p}^{k}\cdot \left(1-p{\right)}^{n-k}$
The binomial coefficient can be written as
$\left(\genfrac{}{}{0}{}{n}{k}\right)=\frac{n\cdot \left(n-1\right)\cdot \dots \cdot \left(n-k+1\right)}{1\cdot 2\cdot \dots \cdot k}$
Step 2
For $n=25$ and $k=5$ the probability is
$P\left(X=5\right)=\frac{25\cdot 24\cdot 23\cdot 22\cdot 21}{1\cdot 2\cdot 3\cdot 4\cdot 5}\cdot {0.2}^{5}\cdot {0.8}^{20}$
This term can be calculated by the most calculators which are allowed in statistic lectures.