Question

According to a​ study, 68​% of all males between the ages of 18 and 24 live at home. ​ (Unmarried college students living in a dorm are counted as living at​ home.) Suppose that a survey is administered and 143 of 206 respondents indicated that they live at home.​ (a) Use the normal approximation to the binomial to approximate the probability that at least 143 respondents live at home.​ (b) Do the results from part​ (a) contradict the​ study?
​(a) ​P(X≥143​)=

Answer 1

Answer:

a) P(X≥143​)=0

b) This contradicts the study as getting a sample with this proportion is almost impossible (if the proportion of 68% is true).

Step-by-step explanation:

If we use the normal approximation to the binomial distribution we have the following parameters (mean and standard deviation):

$\mu=np=0.68*206=140.1\\\\ \sigma=\sqrt{np(1-p)}=\sqrt{206*0.68*0.32}=\sqrt{44.8256}=6.7$

Then, we can calculate the probability of X being equal or more than 143 using the z-score:

$z=\dfrac{X-\mu}{\sigma/\sqrt{n}}=\dfrac{143-140.1}{6.7/\sqrt{206}}=\dfrac{2.9}{0.4668}=6.2124\\\\\\P(x\geq143)=P(z>6.2124)=0$

This contradicts the study as getting a sample with this proportion is almost impossible (if the proportion of 68% is true).