Question

A courier service company wishes to estimate the proportion of people in various states that will use its services. Suppose the true proportion is 0.06. If 373 are sampled, what is the probability that the sample proportion will be less than 0.03

Answer 1

Answer:

$\mu_{p} = p = 0.06$

And the standard error is given by:

$SE_{p} = \sqrt{\frac{\hat p(1-\hat p)}{n}}$

And replacing we got:

$SE_[p]= \sqrt{\frac{0.06*(1-0.06)}{373}}= 0.0123$

And we want to find this probability:

$P(\hat p < 0.03)$

We can calculate the z score for this case and we got:

$z = \frac{\hat p -\mu_p}{\Se_p} = \frac{0.03-0.06}{0.0123}= -2.440$

And using the normal distribution table or excel we got:

$P(\hat p < 0.03)= P(Z$

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

Solution to the problem

For this case we can find the mean and standard error for the sample proportion with these formulas:

$\mu_{p} = p = 0.06$

And the standard error is given by:

$SE_{p} = \sqrt{\frac{\hat p(1-\hat p)}{n}}$

And replacing we got:

$SE_[p]= \sqrt{\frac{0.06*(1-0.06)}{373}}= 0.0123$

And we want to find this probability:

$P(\hat p < 0.03)$

We can calculate the z score for this case and we got:

$z = \frac{\hat p -\mu_p}{\SE_p} = \frac{0.03-0.06}{0.0123}= -2.440$

And using the normal distribution table or excel we got:

$P(\hat p < 0.03)= P(Z$