Answer to Question #96679 in Statistics and Probability for Success

Question

1) If 25% of all students entering a certain university drop out during or at the end of their first year what is the probability that more than 500 of this year's entering class of 1200 will drop out during or at the end of the 1st year
2) a drug manufacturer states that only 7% of the patient using a particular drug will experience side effects a large university hospital uses the drug in treating 300 patients what is the probability that 50 and below of the 300 patients experience side effects?

Answer 1

1) Let $X=$ the number of students who will drop out during or at the end of the 1st year

$$X\sim B(n,p)$$

Given that $p=0.25,n=1200.$ Then

$$np=1200(0.25)=300>10,$$ $$n(1-p)=1200(1-0.25)=900>10$$

We can use a Normal Distribution to approximate a Binomial Distribution

$$X^*\sim N(\mu=np,\sigma^2=np(1-p))$$ $$\mu=np=1200(0.25)=300,$$ $$\sigma^2=np(1-p)=1200(0.25)(1-0.25)=225$$

$X^*\sim N(300,15^2)$

Then

$$Z={X^*-\mu \over \sigma}\sim N(0, 1)$$

Find the probability that more than 500 students will drop out during or at the end of the 1st year 

$$P(X>500)=P\big(Z>{500-300 \over 15}\big)=1-P(Z\leq{40 \over 3})\approx$$ $$\approx7.4\times10^{-41}\approx0$$

2) Let $X=$ the number of patients experiencing side effects.

$$X\sim B(n,p)$$

Given that $p=0.07,n=300.$ Then

$$np=300(0.07)=21\geq10,$$ $$n(1-p)=300(1-0.07)=279\geq10$$

We can use a Normal Distribution to approximate a Binomial Distribution

$$X^*\sim N(\mu=np,\sigma^2=np(1-p))$$ $$\mu=np=300(0.07)=21,$$ $$\sigma^2=np(1-p)=300(0.07)(1-0.07)=19.53$$

$X^*\sim N(21,19.53)$

Then

$$Z={X^*-\mu \over \sigma}\sim N(0, 1)$$

Find the probability that 50 and below of the 300 patients experience side effect

$$P(X\leq50)=P\big(Z\leq{50-21 \over \sqrt{19.53}}\big)\approx$$ $$\approx P(Z\leq6.562161)\approx0.999999999973\approx1$$