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Suppose a plane has 500 seats. From experience, it's known that 10% of people who have bought the ticket do not show up. Now, 550 tickets have been sold for one flight. What is the probability that there will be enough seats for all people who turn up? Hint: Use the normal approximation to the binomial distribution.

My attempt

X ~ B(n, p) and if n is large, then X is approximately N(np, npq)

X ~ B(550,0.9) = N(495,49.5)

B==> P(X<=500) = N==> P(X<500.5)

the output is 0.78 but i am not sure. .. any help?

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Let $X\sim Bin(0.9, 550)$. Now we apply the central limit theorem.

$$P(X\leq 500)\approx\Phi\left( \frac{500-550\cdot 0.9}{\sqrt{550\cdot 0.9\cdot 0.1}} \right)=\Phi\left( 0.710669... \right)\approx\Phi(0.71)$$

  1. Now we can look at a table and see that $\Phi(0.71)=0.76115$. Rounded to two decimal places we get the final result of $\color{blue}{0.76}$.

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  1. Alternatively we can use a calculator and obtain $\Phi(0.710669)=0.761\approx \color{blue}{0.76}$

We can do a better approximation by using the continuity correction factor.

$$P(X\leq 500)\approx\Phi\left( \frac{500+0.5-550\cdot 0.9}{\sqrt{550\cdot 0.9\cdot 0.1}} \right)=\Phi\left( 0.781736... \right)\approx\Phi(0.78)$$

The table gives tells us that $\Phi(0.78)=0.78230\approx 0.78$. This is a better approximation since $P(X\leq 500)=0.7807$ if we do not approximate.

If you are used to calculate with the continuity correction factor at your lecture then $0.78$ should be one of the options.

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