Question

According to the guideline from the Health Department, every over-weight child needs to take a fitness test. Every ten over-weight children would be randomly grouped together to attend the fitness test. According to the past experience, the probability for each child to pass the fitness test is 0.75 and whether a child can pass the test is independent with the test result of other children.
(a) Many test sessions will be conducted regularly, each with 10 independent children. Find the expectation and standard deviation of the number of children can pass the fitness test in a test session.

Answer 1

Answer:

The expected number of children that can pass the fitness test in a test session is 7.5 and the standard deviation is 1.37.

Step-by-step explanation:

For each children, there are only two possible outcomes. Either they pass the health test, or they do not. Whether a child can pass the test is independent with the test result of other children. This means that we use the binomial probability distribution to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

The probability for each child to pass the fitness test is 0.75

This means that $p = 0.75$

10 independent children

This means that $n = 10$ Find the expectation and standard deviation of the number of children can pass the fitness test in a test session.

$E(X) = np = 10*0.75 = 7.5$

$\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{10*0.75*0.25} = 1.37$

The expected number of children that can pass the fitness test in a test session is 7.5 and the standard deviation is 1.37.