Question

assume 85% of people actually show up for their flight on time. because of this, airlines tend to overbook flights. a random sample of 217 booked passengers is taken and whether or not they showed up on time is recorded for each of them.
a.) list the given information with the correct symbols:
select an answer = 0.85
select an answer = 217
b.) can we say that the sampling distribution of \\(\widehat{p}\\) is approximately normal?
?
because \\(np(1 - p)=\\) which is ? 10. round to one decimal place.
c.) what is the probability that at least 90% of them showed up on time? round answer to 4 decimal places.
d.) what is the probability that at most 79% of booked passengers to be on time in a sample of 217? is this unusual? round answer to 4 decimal places.
probability = which select an answer unusual because it is ? 0.05.
submit question

Explanation:

Step1: Identify the symbols

The proportion of people who show up on - time is $p = 0.85$, and the sample size is $n=217$.

Step2: Check the normality of the sampling - distribution

We use the formula $np(1 - p)$ to check if the sampling distribution of $\hat{p}$ is approximately normal. Substitute $n = 217$ and $p=0.85$ into the formula:

$$
LATEXBLOCK0
$$

Since $np(1 - p)=27.7>10$, the sampling distribution of $\hat{p}$ is approximately normal.

Step3: Standardize for part c

We want to find $P(\hat{p}\geq0.90)$. The mean of the sampling distribution of $\hat{p}$ is $\mu_{\hat{p}}=p = 0.85$, and the standard deviation is $\sigma_{\hat{p}}=\sqrt{\frac{p(1 - p)}{n}}=\sqrt{\frac{0.85\times(1 - 0.85)}{217}}=\sqrt{\frac{0.85\times0.15}{217}}=\sqrt{\frac{0.1275}{217}}\approx0.024$.
The z - score is $z=\frac{\hat{p}-\mu_{\hat{p}}}{\sigma_{\hat{p}}}=\frac{0.90 - 0.85}{0.024}=\frac{0.05}{0.024}\approx2.08$.
$P(\hat{p}\geq0.90)=P(Z\geq2.08)=1 - P(Z < 2.08)$. From the standard normal table, $P(Z < 2.08)=0.9812$, so $P(\hat{p}\geq0.90)=1 - 0.9812 = 0.0188$.

Step4: Standardize for part d

We want to find $P(\hat{p}\leq0.79)$. The z - score is $z=\frac{\hat{p}-\mu_{\hat{p}}}{\sigma_{\hat{p}}}=\frac{0.79 - 0.85}{0.024}=\frac{-0.06}{0.024}=-2.5$.
$P(\hat{p}\leq0.79)=P(Z\leq - 2.5)$. From the standard normal table, $P(Z\leq - 2.5)=0.0062$. Since $0.0062<0.05$, it is unusual.

Answer:

a. $p = 0.85$, $n = 217$
b. Yes, $np(1 - p)=27.7>10$
c. $0.0188$
d. $0.0062$, is, $<$