## Related questions with answers

An airline, believing that 5% of passengers fail to show up for flights, overbooks (sells more tickets than there are seats). Suppose a plane will hold 265 passengers, and the airline sells 275 tickets. What’s the probability the airline will not have enough seats, so someone gets bumped?

Solution

VerifiedFor the mean:

$\begin{aligned} E(X) &= \mu = np \\ &= (275)(0.95) \\ &= 261.25 \\ \end{aligned}$

For the standard deviation:

$\begin{aligned} \sigma &= \sqrt{npq} \\ &= \sqrt{(275)(0.95)(1-0.95)} \\ &= 3.614 \\ \end{aligned}$

The z-score:

$\begin{aligned} z &= \dfrac{y-\mu}{\sigma} \\ &= \dfrac{265-261.25}{3.614} \\ &= 1.18 \\ \end{aligned}$

From table Z:

$\begin{aligned} P(X>265) &= P(Z>1.18) \\ &= 1-P(Z<1.18) \\ &= 1-0.8810 \\ &= \boxed{0.119} \\ \end{aligned}$

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