Probability that out of their next 100 free throws, they will make between $75$ and $80$, inclusive in basketball game.

Question

State Tech's basketball team, the Fighting Logarithms, have 70% foul-shooting percentage.

(a) Write a formula for the exact probability that out of their next 100 free throws, they will make between $75$ and $80$, inclusive.

Solution: Let X = the number they will make.

Then the formula for probability that out of their next $100$ free throws, they will make between $75$ and $80$, inclusive can be found by the general formula.

Then by formula if $X$ is a binomial random variable with parameters $n$ and $p$ then $P(a \leq X \leq b) = F_{Z} [\frac{b + 0.5 - np}{\sqrt{np(1-p)}}] - F_{Z} [\frac{a - 0.5 - np}{\sqrt{np(1-p)}}]$.

Then finding $np = 100*0.70 = 70$ and $\sqrt{np(1-p) =} \sqrt{70(1-0.70)} = \sqrt{21}$

Then plugging we have $P(75 \leq X \leq 80) = F_{Z}[\frac{80 + 0.5 - 70}{\sqrt21}] - F_{Z}[\frac{75 - 0.5 - 70}{\sqrt21}]$.

b) Approximate the probability asked for in part (a).

Solution: Using our formula and the appendix/table for cumulative areas under the standard normal distributions we have, $P(75 \leq X \leq 80) = F_{Z}[\frac{80 + 0.5 - 70}{\sqrt21}] - F_{Z}[\frac{75 - 0.5 - 70}{\sqrt21}] = F_{Z}[\frac{10.5}{\sqrt21}] - F_{Z}[\frac{4.5}{\sqrt21}] = F_{Z}[2.2912] - F_{Z}[0.98198] = 0.9890 - 0.8365 = 0.1525$.

Can anyone please verify this? Any feedback/help would be helpful. Thank you.

Answer 1

What you have calculated for a and b is the solution for b. This is o.k.

To get the formula for the excact probability, you have to use the binomial distribution:

$$P(75 \leq X \leq 80)=\sum_{k=75}^{80} {100 \choose k} 0.7^k\cdot 0.3^{100-k}$$

$P(75 \leq X \leq 80) \color{red} = F_{Z}[\frac{80 + 0.5 - > 70}{\sqrt21}] - F_{Z}[\frac{75 - 0.5 - 70}{\sqrt21} ]= \ldots$

It is better to use the $\approx$ - sign instead of the first equality-sign. As you know, it is an approximation.