Answer:
The mean and the standard deviation of the number of students with laptops are 1.11 and 0.836 respectively.
Step-by-step explanation:
Let X = number of students who have laptops.
The probability of a student having a laptop is, P (X) = p = 0.37.
A random sample of n = 30 students is selected.
The event of a student having a laptop is independent of the other students.
The random variable X follows a Binomial distribution with parameters n and p.
The mean and standard deviation of a binomial random variable X are:
[tex]\mu=np\\\sigma=\sqrt{np(1-p)}[/tex]
Compute the mean of the random variable X as follows:
[tex]\mu=np=30\times0.37=1.11[/tex]
The mean of the random variable X is 1.11.
Compute the standard deviation of the random variable X as follows:
[tex]\sigma=\sqrt{np(1-p)}=\sqrt{30\times0.37\times(1-0.37)}=\sqrt{0.6993}=0.836[/tex]
The standard deviation of the random variable X is 0.836.
