Question

An article reported that in a large study carried out in the state of New York, approximately 60% of the study subjects lived within 1 mile of a hazardous waste site. Let p denote the proportion of all New York residents who live within 1 mile of such a site, and suppose that p = 0.6.(c) When n = 400, what is P(0.55 ? p? ? 0.65)?

Answer 1

Answer:

95.86%.

Step-by-step explanation:

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.

In this problem, we have that:

$\mu = 0.6, \sigma = \sqrt{\frac{p(1-p)}{n}} = 0.0245$

$P(0.55 \leq x \leq 0.65$

This probability is the pvalue of Z when $X = 0.65$ subtracted by the pvalue of Z when $X = 0.55$. So

X = 0.65

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

$Z = \frac{0.65 - 0.6}{0.0245}$

$Z = 2.04$

$Z = 2.04$ has a pvalue of 0.9793

X = 0.55

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

$Z = \frac{0.55 - 0.6}{0.0245}$

$Z = -2.04$

$Z = -2.04$ has a pvalue of 0.0207

So this probability is 0.9793 - 0.0207 = 0.9586 = 95.86%.