Question

find the mean, variance, and standard deviation of the binomial distribution with the given values of n and p. n = 124, p = 0.69 the mean, μ, is . (round to the nearest tenth as needed.)

Explanation:

Step1: Recall mean formula

The mean $\mu$ of a binomial distribution is given by $\mu = np$.
$\mu=124\times0.69$

Step2: Calculate the mean

$\mu = 124\times0.69 = 85.56\approx85.6$

Step3: Recall variance formula

The variance $\sigma^{2}$ of a binomial distribution is $\sigma^{2}=np(1 - p)$.
$\sigma^{2}=124\times0.69\times(1 - 0.69)$

Step4: Calculate the variance

$\sigma^{2}=124\times0.69\times0.31=124\times0.2139 = 26.5236$

Step5: Recall standard - deviation formula

The standard deviation $\sigma$ is the square - root of the variance, $\sigma=\sqrt{\sigma^{2}}$.
$\sigma=\sqrt{26.5236}\approx5.15$

Answer:

The mean $\mu$ is $85.6$, the variance $\sigma^{2}$ is $26.5$ (rounded to one decimal place), and the standard deviation $\sigma$ is $5.2$ (rounded to one decimal place).