A sell-out crowd of 42,200 is expected at Cleveland's Jacobs Field for next Tuesday's game with the Baltimore Orioles, the last before a long road trip. The ballpark's records from games played either in the season, she knows that, on the average, 38% of all those in attendance will buy a hot dog. How large an order should she place if she wants to have no more that a 20% chance of demand exceeding supply?.
Attempt:
Let $X$ = amount of order. Here X is a binomial distribution with $p = 0.38$ and $n = 42,200$.
Then
$P(X > x) = 0.20$ $\rightarrow$ $P(X \leq x) = 0.80$.
Then from the continuity correction we have
$P(X \leq x)$ = $P(\frac{(x - np)}{\sqrt{np(1-p)}} \leq \frac{(X - 42,200(0.38))}{\sqrt{42,200(0.38)(1-0.38)}}) = 0.80$
Then putting $P( z \leq \frac{(x - 42,200(0.38))}{\sqrt{42,200(0.38)(1-0.38)}}) = P(z \leq \frac{x - 16036}{99.7111}) = 0.80$
I know I have to use the standard normal table to determine where $\frac{x - 16036}{99.7111}$ is about the standard normal table, then solve for $x$. But I don't know how to continue. Can anyone please help? Any feedback/help can help. Thank you.
First of all, define a normalized variable $$ Z={X-\mu\over \sigma} $$ Then, on the normal distribution table, you need to look for a probability equal to about 0.8. The linked table gives $z_0\approx0.84$ so we have $$ P(Z<0.84)\approx0.80 $$ Thus to get the right number of hotdogs, $$ {x_0-16036\over100}=z_0=0.84 $$ so $$ x=16036+84=16120 $$