What are the mean and standard deviation of a binomial probability distribution with n=150 and...

mean =np
${\displaystyle s\tan darddeviation=\sqrt{npq}}$
Explanation:
${\displaystyle mean=150\times \frac{7}{9}=\frac{350}{3}\approx 116.7}$
${\displaystyle s\tan darddeviation=\sqrt{150\times \frac{7}{9}\times (1-\frac{7}{9})}\approx 5.09}$

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
${\displaystyle \mu =150\cdot \frac{7}{9}}$
Multiplying the expression we have
${\displaystyle \mu =115.5}$
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 umber of trials and p is the probability of success on a single trial. Substituting values fo this problem, we have
${\displaystyle \sigma =\sqrt{150\cdot \frac{7}{9}\cdot (1-\frac{7}{9})}}$
Evaluating the expression on the right, we have
${\displaystyle \sigma =\sqrt{26.565}}$
${\displaystyle \sigma =5.09}$