A college basketball player makes 80% of his freethrows. Over the course of the season he will attempt 100 freethrows. Assuming free throw attempts are independent, the probability that the number of free throws he makes exceeds 80 is approximately:. A) 0.2000
B) 0.2266
C) 0.5000
D) 0.7734

The probability that the number of free throws he makes exceeds 80 is approximately 0.50

Step-by-step explanation:

According to the given data we have the following:

P(Make a Throw) = 0.80%

n=100

Binomial distribution:

mean:   np = 0.80*100= 80

hence, standard deviation=√np(1-p)=√80*0.20=4

Therefore, to calculate the probability that the number of free throws he makes exceeds 80 we would have to make the following calculation:

P(X>80)= 1- P(X<80)

You could calculate this value via a normal :

P(Z<(80-80)/4)=1-P(Z<0)=1-50=0.50

The probability that the number of free throws he makes exceeds 80 is approximately 0.50

answer: i don't understand. what are the options

step-by-step explanation:

3/6 chance

step-by-step explanation:

half of the nubers are odd numbers (1,3,5) which in that case if you would be told to simplify it would be 1/2