What are the mean and standard deviation of a binomial probability distribution with n=17 and...

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 ${\displaystyle p=\frac{18}{32}=\frac{9}{16}}$ the mean is
${\displaystyle 17\cdot \frac{9}{16}=\frac{153}{16}=9.5}$ and variance is ${\displaystyle 17\cdot \frac{9}{16}\cdot (1-\frac{9}{16})=\frac{1071}{256}}$. Standard deviation is the square root of variance so we have ${\displaystyle \sqrt{\frac{1071}{256}}=3\frac{\sqrt{119}}{16}\approx 2.0454}$.

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
${\displaystyle \mu =n\cdot p}$
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
${\displaystyle \mu =17\cdot \frac{18}{32}}$
Multiplying the expression we have
${\displaystyle \mu =9.52}$
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
${\displaystyle \sigma =\sqrt{n\cdot p\cdot (1-p)}}$
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
${\displaystyle \sigma =\sqrt{17\cdot \frac{18}{32}\cdot (1-\frac{18}{32})}}$
Evaluating the expression on the right, we have
${\displaystyle \sigma =\sqrt{4.1888}}$
${\displaystyle \sigma =2.0454}$