 zakinutuzi

2022-01-16

What are the mean and standard deviation of a binomial probability distribution with n=17 and $p=\frac{18}{32}$? Kirsten Davis

Expert

Explanation:
Unless you have to calculate everything each time, we could simply use known formulas.
In binomial distribution the mean is given by np and variance by np(1-p).
Since in our case n=17 and $p=\frac{18}{32}=\frac{9}{16}$ the mean is
$17\cdot \frac{9}{16}=\frac{153}{16}=9.5$ and variance is $17\cdot \frac{9}{16}\cdot \left(1-\frac{9}{16}\right)=\frac{1071}{256}$. Standard deviation is the square root of variance so we have $\sqrt{\frac{1071}{256}}=3\frac{\sqrt{119}}{16}\approx 2.0454$. ramirezhereva

Expert

Solution:
The mean of the binomial distribution is interpreted as the mean number of successes for the distribution. To find the mean, use the formula
$\mu =n\cdot p$
where n is the number of trials and p is the probability of success on a single trial. Substituting values for this problem, we have
$\mu =17\cdot \frac{18}{32}$
Multiplying the expression we have
$\mu =9.52$
The standard deviation of the binomial distribution is interpreted as the standard deviation of the number of successes for the distribution. To find the standard deviation, use the formula
$\sigma =\sqrt{n\cdot p\cdot \left(1-p\right)}$
where n is the number of trials and p is the probability of success on a single trial. Substituting values fo this problem, we have
$\sigma =\sqrt{17\cdot \frac{18}{32}\cdot \left(1-\frac{18}{32}\right)}$
Evaluating the expression on the right, we have
$\sigma =\sqrt{4.1888}$
$\sigma =2.0454$

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