Becky made 2 video tapes while a giant hummingbird fed 4 times and a ruby throated hummingbird fed 8 times from her new feeder.
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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 <em>X</em> = number of students who have laptops.
The probability of a student having a laptop is, P (X) = <em>p</em> = 0.37.
A random sample of <em>n</em> = 30 students is selected.
The event of a student having a laptop is independent of the other students.
The random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> and <em>p</em>.
The mean and standard deviation of a binomial random variable <em>X</em> are:
Compute the mean of the random variable <em>X</em> as follows:
The mean of the random variable <em>X</em> is 1.11.
Compute the standard deviation of the random variable <em>X</em> as follows:
The standard deviation of the random variable <em>X</em> is 0.836.