Question

An IAB study on the state of original digital video showed that original digital video is becoming increasingly popular. Original digital video is defined as professionally produced video intended only for ad-supported online distribution and viewing. According to IAB data, 30% of American adults 18 or older watch original digital videos each month.
1. Suppose that you take a sample of 100 U.S. adults, what is the probability that fewer than 25 in your sample will watch original digital videos?

Answer 1

Answer:

11.51% probability that fewer than 25 in your sample will watch original digital videos

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:

$n = 100, p = 0.30$

So

$\mu = E(X) = np = 100*0.3 = 30$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{100*0.3*0.7} = 4.58$

What is the probability that fewer than 25 in your sample will watch original digital videos?

Using continuity correction, this is $P(X < 25 - 0.5) = P(X < 24.5)$, which is the pvalue of Z when X = 24.5. So

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

$Z = \frac{24.5 - 30}{4.58}$

$Z = -1.2$

$Z = -1.2$ has a pvalue of 0.1151

11.51% probability that fewer than 25 in your sample will watch original digital videos