Question

Assume that a procedure yields a binomial distribution with n trials and the probability of success for one trial is p. Use the given values of n and p to find the mean mu and standard deviation sigma. Also, use the range rule of thumb to find the minimum usual value mu minus 2 sigma and the maximum usual value mu plus 2 sigma.

n=1580,=1/4

Answer 1

Concepts and reason

The concept used of Binomial distribution and basic mean-standard deviation manipulation is used here.

The binomial distribution is a probability distribution of each trial which can result in either success or failure. The mean of the binomial distribution can be obtained from the number of trials and the probability of success in each trial. Similarly, the standard deviation of the binomial distribution can be obtained by square root of the variance.

Fundamentals

Suppose there is a Binomial distribution with $n$independent trials and the probability of success and probability of failure $p$and$q$respectively. Then, the mean and standard deviation can be obtained as follows:

$Mean$ = $np$

$stdev = \sqrt {np\left( {1 - p} \right)}$

The range of two standard deviation away from the mean can be obtained by using the formulae:

$\begin{array}{l}\\R1 = Mean + 2.stddev\\\\R2 = Mean - 2.stddev\\\end{array}$

The following information has been provided about the binomial distribution:

$\begin{array}{l}\\n = 1580\\\\p = \frac{1}{4}\\\end{array}$

The mean can be obtained as follows:

$\begin{array}{c}\\Mean = np\\\\ = 1580\left( {\frac{1}{4}} \right)\\\\ = 395\\\end{array}$

Similarly, the standard deviation can be obtained as follows:

$\begin{array}{c}\\stdev = \sqrt {np\left( {1 - p} \right)} \\\\ = \sqrt {1580\left( {\frac{1}{4}\left( {\frac{3}{4}} \right)} \right)} \\\\ = 17.212\\\end{array}$

The range of two standard deviation can be obtained as:

$\begin{array}{c}\\R1 = Mean + 2.stddev\\\\ = 395 + 2.\left( {17.212} \right)\\\\R2 = Mean - 2.stddev\\\\ = 395 - 2.\left( {17.212} \right)\\\end{array}$

The two expressions after solving would yield the following two values:

$\begin{array}{l}\\R1 = 360.576\\\\R2 = 429.424\\\end{array}$

Ans:

The range of two standard deviation away from the mean has been obtained as$\left( {360.576,429.424} \right)$.