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There are 90 people in the restaurant. The probability of someone ordering a drink with food is 60%. Use Normal approximation of Binomial Distribution to answer the following 6 questions.

  1. What is the mean of the Normal distribution?
  2. What is the standard deviation of the normal distribution?
  3. What is the probability that exactly 50 people will order a drink?
  4. What the probability that more than 50 people will order a drink?
  5. What is the probability that less than 50 people will order a drink?
  6. What is the probability that between 52 or more and 56 or less people will order a drink?

1 and 2 I get (54 and 4.64758, respectively). 3 - 6 not so much, especially how to look up z values and probabilities in the standard normal table. I only have access to the 0 to z normal distribution table, like page one here:

http://chemeng.iisc.ac.in/venu/tables.pdf

Thanks.

cyclops
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    Please express your gratitude (if there's any) via the way of upvoting or clicking the accept (green check mark) for the answer for this and your previous questions. – callculus42 Oct 10 '18 at 20:56

1 Answers1

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You have to standardize the random variable. Let $X$ the binomial distributed random variable then

$$P(X\leq x)\approx\Phi\left( \frac{x+0.5-\mu}{\sigma} \right)$$

with $\mu=90\cdot 0.6=54$ and $\sigma=\sqrt{90\cdot 0.6\cdot 0.4}\approx 4.64758$. It confirms your results.

$\Phi(z)$ is the cdf of the standard normal distribution and $+0.5$ is the continuity correction factor. Now some hints. At $5)$ I demonstrate how to evaluate $P(X\leq x)$.

3) $P(X=x)=P(X\leq x)-P(X\leq x-1)$ In your case $x=50$ and $x-1=49$

4) $P(X> 50)=1-P(X\leq 50)$

5) $P(X< 50)=P(X\leq 49)$

$P(X\leq 49)\approx \Phi\left( \frac{49+0.5-54}{4.64758} \right)=\Phi\left( -0.97\right)=1-\Phi\left( 0.97 \right)$

Now you look at the table and look for the probability, where $\Phi(0.97)=p$ This is at the row with z=0.9 and the column with $0.07$. The value is $0.3340$ and we have to add $0.5$ since $P(Z<0)=0.5$: 0.834

Thus $P(X<50)\approx 1-0.834=0.166$

Using the binomial distribution confirms the approximation:

$$\sum_{x=0}^{49} {{90} \choose x} 0.6^x\cdot 0.4^{90-x}=0.1663$$

6) $P(52\leq X\leq 56)=P(X\leq 56)-P(X\leq 51)$

callculus42
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