Question

find the mean, variance, and standard deviation of the binomial distribution with the given values of n and p.

\\(n = 90\\), \\(p = 0.9\\)

the mean, \\(\mu\\), is 81. (round to the nearest tenth as needed.)

the variance, \\(\sigma^2\\), is 8.1. (round to the nearest tenth as needed.)

the standard deviation, \\(\sigma\\), is . (round to the nearest tenth as needed.)

Explanation:

Calculate the mean

$$ \mu = n \cdot p = 90 \cdot 0.9 = 81 $$

Calculate the variance

$$ \sigma^2 = n \cdot p \cdot (1 - p) = 90 \cdot 0.9 \cdot 0.1 = 8.1 $$

Calculate the standard deviation

$$ \sigma = \sqrt{\sigma^2} = \sqrt{8.1} \approx 2.846 \approx 2.8 $$

Answer:

Find the mean, variance, and standard deviation of the binomial distribution with the given values of n and p.
\(n = 90, p = 0.9\)

The mean, \(\mu\), is <blank>81</blank>. (Round to the nearest tenth as needed.)

The variance, \(\sigma^2\), is <blank>8.1</blank>. (Round to the nearest tenth as needed.)

The standard deviation, \(\sigma\), is <blank>2.8</blank>. (Round to the nearest tenth as needed.)