Question

A Food Marketing Institute found that 52% of households spend more than $125 a week on groceries. Assume the population proportion is 0.52 and a simple random sample of 99 households is selected from the population. What is the probability that the sample proportion of households spending more than $125 a week is more than than 0.38?

Answer 1

Answer:

$P(p>0.38) = P(z>-2.79)$

And using the complement rule we got:

$P(p>0.38) = P(z>-2.79)=1-P(Z$

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".  

Solution to the problem

For this case wwe know that p = 0.52 and n = 99 and we can check if we can use the normal approximation for the proportion distribution.

$np= 99*0.52= 51.48>10$

$n(1-p) =99*(1-0.52) = 47.52>10$

So then we can use the normal approximation.

The population proportion have the following distribution

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

The mean is given by:

$\mu_{p} = 0.52$

And the standard error is given by:

$\sigma_{p} = \sqrt{\frac{0.52*(1-0.52)}{99}}= 0.0502$

We want to calculate this probability:

$P(p >0.38)$

And for this case we can calculate the z score given by:

$z = \frac{p \mu_p}{\sigma_p}$

And replacing we got:

$z=\frac{0.38-0.52}{0.0502} = -2.79$

And using this formula:

$P(p>0.38) = P(z>-2.79)$

And using the complement rule we got:

$P(p>0.38) = P(z>-2.79)=1-P(Z$