Question

Eighty percent of the students graduating from high school in a small Iowa farm town attend college. The town's chamber of commerce randomly selects 45 recent graduates and inquires whether or not they will attend college. Find the probability that at least 82% of the surveyed students will be attending college.

Answer 1

Answer: 0.6368

Step-by-step explanation:

Given : The proportion of z-test for population mean : $\hat{p}= 0.80$

Sample size : $n=45$

We assume that this is normally distributed.

The test statistic for population proportion :-

$z=\dfrac{\hat{p}-p_0}{\sqrt{\dfrac{p_0(1-p_0)}{n}}}$

Now, for $p_0=0.82$, we have

$z=\dfrac{0.80-0.82}{\sqrt{\dfrac{0.82(1-0.82)}{45}}}\\\\\approx-0.35$

By using the standard normal distribution table , the probability that at least 82% of the surveyed students will be attending college will be :-

$P(p\geq0.82)=P(z\geq-0.35)\\\\=1-P(z$

Hence, the probability that at least 82% of the surveyed students will be attending college = 0.6368