Expert Answer to "What are the mean and standard deviation of..."

Kaspaueru2

Answered question

2022-01-17

What are the mean and standard deviation of a binomial probability distribution with n=18 and

p = \frac{2}{32}

Answer & Explanation

Marcus Herman

Beginner2022-01-17Added 41 answers

Explanation:
A binomial random variable X counts the number of successes in n trials, when each trial is independent from the others with a constant probability p of success.
The mean of X is

μ = n \cdot p

and the standard deviation of X is

σ = \sqrt{n \cdot p \cdot (1 - p)} .

The mean is

μ = n \cdot p = 18 \cdot \frac{1}{16} = \frac{9}{8}

and the standard deviation is

σ = \sqrt{n \cdot p \cdot (1 - p)} = \sqrt{\frac{135}{128}} = \frac{3 \sqrt{15}}{8 \sqrt{2}} = \frac{3 \sqrt{30}}{16} \approx 1.027

Annie Gonzalez

Beginner2022-01-18Added 41 answers

The formula for the population mean is

μ = n \cdot p

, so therefore we get:

μ = n \cdot p

= 18 \cdot \frac{2}{32}

=1.125
On the other hand, the variance is computed using the following formula:

σ^{2} = n p (1 - p)

. Therefore, the calculation goes like:

σ^{2} = n p (1 - p)

= 18 \cdot \frac{2}{32} (1 - \frac{2}{32})

=1.05
And finally, taking square root to the variance we get that the population standard deviation is:

σ = \sqrt{σ^{2}} = \sqrt{1.05} = 1.027

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