Question

question 12 assume that a procedure yields a binomial distribution with n = 1227 trials and the probability of success for one trial is p = 0.84. find the mean for this binomial distribution. (round answer to one decimal place.) μ = find the standard deviation for this distribution. (round answer to two decimal places.) σ = 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 =

Explanation:

Step1: Recall mean formula

The mean $\mu$ of a binomial distribution is given by $\mu = np$. Here, $n = 1227$ and $p=0.84$.
$\mu=1227\times0.84$

Step2: Calculate the mean

$\mu = 1227\times0.84=1030.68\approx1030.7$

Step3: Recall standard - deviation formula

The standard deviation $\sigma$ of a binomial distribution is $\sigma=\sqrt{np(1 - p)}$. Substitute $n = 1227$ and $p = 0.84$. First, find $1-p=1 - 0.84=0.16$.
$\sigma=\sqrt{1227\times0.84\times0.16}$

Step4: Calculate the standard deviation

$\sigma=\sqrt{1030.68\times0.16}=\sqrt{164.9088}\approx12.84$

Step5: Find minimum and maximum usual values

The range - rule of thumb for usual values is $\mu\pm2\sigma$.
Minimum usual value: $\mu - 2\sigma=1030.7-2\times12.84=1030.7 - 25.68 = 1005.02\approx1005$
Maximum usual value: $\mu + 2\sigma=1030.7+2\times12.84=1030.7 + 25.68 = 1056.38\approx1056$
The interval of usual values is $[1005,1056]$.

Answer:

$\mu = 1030.7$
$\sigma\approx12.84$
Usual values: $[1005,1056]$