Answer to Question #98592 in Statistics and Probability for natasha

Question

Because many passengers who make reservations do not show up, airlines often
overbook flights (sell more tickets than there are seats). A Boeing 767-400ER holds 245
passengers. The airline believes the rate of passenger not showing up is 5% and sells 255
tickets. [10 Marks]
a. Use Normal approximation to determine the binomial probability of at least 246
passengers showing up. [9 Marks]
b. Should the airline change the number of tickets it sells for the flight? Explain. [1
Mark]

Answer 1

Let $X=$ the number of customers who prefer to buy items that they have seen advertised on television: $X\sim B(n, p)$

Given that $p=0.95, n=255$

The mean value and standard deviation of a binomial random variable $X$ are

$$\mu=np, \sigma=\sqrt{np(1-p)},$$

respectively.

$$np=255(0.95)=242.25>10$$ $$n(1-p)=255(0.05)=12.75>10$$

We can use Normal approximation to the Binomial

$X$ has approximately a normal distribution with $\mu=np$ and $\sigma=\sqrt{np(1-p)}: X\sim N(\mu, \sigma^2)$

Then

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

While the normal distribution is continuous (it includes all real numbers), the binomial distribution can only take integers. The small correction is an allowance for the fact that we’re using a continuous distribution.

Continuity Correction Factor Table

$$\begin{array}{c:c} \text{Discrete} & \text{Continuous} \\ \hline X=n & n-0.5<X<n+0.5 \\ \hline X>n & X>n+0.5 \\\hline X\leq n & X<n+0.5 \\\hline X<n & X<n-0.5 \\ \hdashline X\geq n & X>n-0.5 \end{array}$$

$$\mu=np=255(0.95)=242.25$$

$$\sigma=\sqrt{np(1-p)}=\sqrt{255(0.95)(1-0.95)}\approx3.4803$$

a. Use Normal approximation to determine the binomial probability of at least 246 passengers showing up.

$$P(X>246-0.5)=1-P(X\leq245.5)=$$

$$=1-P(Z\leq{245.5-242.25 \over 3.4803})\approx1-P(Z\leq0.933828)\approx$$

$$\approx1-P(Z\leq0.933828)\approx1-0.82480357\approx$$

$$\approx0.1752 (17.52 \%)$$

The binomial probability of at least 246 passengers showing up is $0.1752$

(b) Since $17.52 \%>5\%,$ then the airline change (decrease) the number of tickets it sells for the flight.