Question

According to a 2009 Reader's Digest article, people throw away approximately 11% of what they buy at the grocery store. Assume this is the true proportion and you plan to randomly survey 108 grocery shoppers to investigate their behavior. What is the probability that the sample proportion exceeds 0.2

Answer 1

Answer:

0.14% probability that the sample proportion exceeds 0.2

Step-by-step explanation:

I am going to use the binomial approximation to the normal 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:

$n = 108, p = 0.11$

So

$\mu = E(X) = np = 108*0.11 = 11.88$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{108*0.11*0.89} = 3.2516$

What is the probability that the sample proportion exceeds 0.2

This is 1 subtracted by the pvalue of Z when X = 0.2*108 = 21.6. So

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

$Z = \frac{21.6 - 11.8}{3.2516}$

$Z = 2.99$

$Z = 2.99$ has a pvalue of 0.9986

1 - 0.9986 = 0.0014

0.14% probability that the sample proportion exceeds 0.2