Question

It’s estimated that 52% of American adults have incurred credit card debt. The department of finance surveys 2500 adults for a report.

Determine the probability that between 1200 and 1450 of those surveyed incurred debt.

Answer 1

Answer:

100% probability that between 1200 and 1450 of those surveyed incurred debt.

Step-by-step explanation:

We 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:

$n = 2500, p = 0.52$

So

$\mu = E(X) = 2500*0.52 = 1300$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{2500*0.52*0.48} = 24.98$

Determine the probability that between 1200 and 1450 of those surveyed incurred debt.

This is the pvalue of Z when X = 1450 subtracted by the pvalue of Z when X = 1200. So

X = 1450

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

$Z = \frac{1450 - 1300}{24.98}$

$Z = 6$

$Z = 6$ has a pvalue of 1

X = 1200

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

$Z = \frac{1200 - 1300}{24.98}$

$Z = -4$

$Z = -4$ has a pvalue of 0

1 - 0 = 1

100% probability that between 1200 and 1450 of those surveyed incurred debt.