Question

the probability that an individual is left - handed is 0.12. in a class of 40 students, what is the mean and standard deviation of the number of left - handers in the class?

mean: 40; standard deviation: 2.29
mean: 40; standard deviation: 2.06
mean: 4.8; standard deviation: 2.06
mean: 4.8; standard deviation: 2.29

Explanation:

Step1: Identify the distribution

This is a binomial - distribution problem with $n = 40$ (number of trials, i.e., number of students) and $p=0.12$ (probability of success, i.e., probability of a student being left - handed).

Step2: Calculate the mean

The mean of a binomial distribution is given by $\mu=np$. So, $\mu = 40\times0.12=4.8$.

Step3: Calculate the standard deviation

The standard deviation of a binomial distribution is $\sigma=\sqrt{np(1 - p)}$. Substitute $n = 40$ and $p = 0.12$ into the formula: $1-p=1 - 0.12 = 0.88$, then $\sigma=\sqrt{40\times0.12\times0.88}=\sqrt{4.224}\approx2.06$.

Answer:

mean: 4.8; standard deviation: 2.06