Use the appropriate normal distribution to approximate the resulting binomial distribution. Preview Showcase, a research firm, screens pilots of new TV shows before a randomly selected audience and then solicits their opinions of the shows. Based on past experience, 20% of those who get complimentary tickets are "no-shows." The theater has a seating capacity of 450. Management has decided, for public relations reasons, that a person who has been solicited for a screening should have a probability of .99 of being seated. How many tickets should the company send out to prospective viewers for each screening?
A) 590 tickets the company should send out.
B) 535 tickets the company should send out.
C) 549 tickets the company should send out.
D) 490 tickets the company should send out.

Question

Quizplus · Accepted Answer

To ensure a 99% chance of seating, we use the normal approximation to the binomial distribution. Let $X$ be the number of people who show up. $X$ follows approximately a normal distribution with mean $\mu = np$ and standard deviation $\sigma = \sqrt{np(1-p)}$, where $n$ is the number of tickets sent, $p = 0.8$ (since 20% are no-shows, 80% are expected to show up), and we want $P(X \leq 450) = 0.99$.Using the standard normal distribution, we find the z-score corresponding to 0.99 probability, which is approximately 2.33. We then set up the equation for the z-score: $$2.33 = \frac{450 - 0.8n}{\sqrt{0.8 \cdot 0.2 \cdot n}}$$Solving this equation for $n$ gives us the number of tickets to send out to achieve the desired probability. The correct calculation leads to a number close to 535, which means 535 tickets should be sent out to ensure a 99% chance of seating for everyone who shows up.

Use the Appropriate Normal Distribution to Approximate the Resulting Binomial