Answer to Question #96691 in Statistics and Probability for Gisel

Question

Last year there were 879 challenges made to referee calls in professional tennis singles play. Among those challenges, 231 were upheld. Historically,25% of calls are upheld. Find the probable that among the 879 challenges last year 231 or more were upheld

Answer 1

Let $X=$ the number of upheld challenges: $X\sim B(n,p).$

Given that $p=0.25,n=879.$ Then

$$np=879(0.25)=219.75\geq10$$ $$n(1-p)=879(1-0.25)=659.25\geq10$$

We can use a Normal Distribution to approximate a Binomial Distribution

$$X^*\sim N(\mu=np,\sigma^2=np(1-p))$$ $$\mu=np=879(0.25)=219.75$$ $$\sigma^2=np(1-p)=879(0.25)(1-0.25)=164.8125$$

$X^*\sim N(219.75,164.8125)$

Then

$$Z={X^*-\mu \over \sigma}\sim N(0,1)$$

Find the probability that among the 879 challenges last year 231 or more were upheld

$$P(X^*\geq231)=P\big(Z\geq{231-219.75 \over\sqrt{164.8125}}\big)=$$

$$=1-P\big(Z<{231-219.75 \over\sqrt{164.8125}}\big)\approx1-0.80956906=$$

$$=0.19043094$$