Question #98592
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]
1
Expert's answer
2019-11-13T11:54:57-0500

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

Given that p=0.95,n=255p=0.95, n=255

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


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

respectively.


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

We can use Normal approximation to the Binomial

XX has approximately a normal distribution with μ=np\mu=np and σ=np(1p):XN(μ,σ2)\sigma=\sqrt{np(1-p)}: X\sim N(\mu, \sigma^2)

Then


Z=XμσN(0,1)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

DiscreteContinuousX=nn0.5<X<n+0.5X>nX>n+0.5XnX<n+0.5X<nX<n0.5XnX>n0.5\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}

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

σ=np(1p)=255(0.95)(10.95)3.4803\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>2460.5)=1P(X245.5)=P(X>246-0.5)=1-P(X\leq245.5)=

=1P(Z245.5242.253.4803)1P(Z0.933828)=1-P(Z\leq{245.5-242.25 \over 3.4803})\approx1-P(Z\leq0.933828)\approx

1P(Z0.933828)10.82480357\approx1-P(Z\leq0.933828)\approx1-0.82480357\approx


0.1752(17.52%)\approx0.1752 (17.52 \%)

The binomial probability of at least 246 passengers showing up is 0.17520.1752


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



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