Question

assume that a procedure yields a binomial distribution with n = 821 trials and the probability of success for one trial is p = 0.21. find the mean for this binomial distribution. (round answer to one decimal place.) μ = blank find the standard deviation for this distribution. (round answer to two decimal places.) σ = blank use the range - rule of thumb to find the minimum usual value μ - 2σ and the maximum usual value μ + 2σ. enter answer as an interval using square - brackets only with whole numbers. usual values = blank

Explanation:

Step1: Calculate the mean

The formula for the mean of a binomial distribution is $\mu = np$. Given $n = 821$ and $p=0.21$, we have $\mu=821\times0.21$.
$\mu = 821\times0.21=172.41\approx172.4$

Step2: Calculate the standard - deviation

The formula for the standard deviation of a binomial distribution is $\sigma=\sqrt{np(1 - p)}$. Substitute $n = 821$ and $p = 0.21$ into the formula. First, find $1-p=1 - 0.21 = 0.79$. Then $np(1 - p)=821\times0.21\times0.79 = 172.41\times0.79 = 136.2039$. So, $\sigma=\sqrt{136.2039}\approx11.67$.

Step3: Find the minimum and maximum usual values

The range - rule of thumb states that the minimum usual value is $\mu - 2\sigma$ and the maximum usual value is $\mu+2\sigma$.
$\mu - 2\sigma=172.41-2\times11.67=172.41 - 23.34 = 149.07\approx149$
$\mu+2\sigma=172.41 + 2\times11.67=172.41+23.34 = 195.75\approx196$
The interval of usual values is $[149,196]$.

Answer:

$\mu = 172.4$
$\sigma\approx11.67$
usual values = $[149,196]$