Question

20% of US High School teens vape. A local High School has implemented campaigns to reduce vaping among students and believes that the percentage of students who vape at this High School is lower than the national average. School administration implements a survey of 300 randomly selected students. They find that 51 of the 300 vape.
If the true proportion of students who vape at this school is 20%, what is the approximate probability of observing 51 or fewer vapers in a random sample of 300?

Answer 1

Answer:

10.93% probability of observing 51 or fewer vapers in a random sample of 300

Step-by-step explanation:

I am going to use the normal approximation to the binomial to solve this question.

Binomial probability distribution

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

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

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

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$, $\sigma = \sqrt{V(X)}$.

In this problem, we have that:

$n = 300, p = 0.2$

So

$\mu = E(X) = np = 300*0.2 = 60$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{300*0.2*0.8} = 6.9282$

What is the approximate probability of observing 51 or fewer vapers in a random sample of 300?

Using continuity corrections, this is $P(X \leq 51 + 0.5) = P(X \leq 51.5)$, which is the pvalue of Z when X = 51.5 So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{51.5 - 60}{6.9282}$

$Z = -1.23$

$Z = -1.23$ has a pvalue of 0.1093.

10.93% probability of observing 51 or fewer vapers in a random sample of 300