LeBron James has a freethrow success percentage of 74 Assume

LeBron James has a free-throw success percentage of 74%. Assume the free-throw shots are independent; that is, success or failure on one shot does not affect the chance of success on another shot. Let r be a random variable that represents each free-throw attempt. Find the probability distribution for r if he has 600 free throw attempts during this season.

a. Find the mean and standard deviation

b. Find the probability that he makes at least 475 free-throws this season. P(r ) = P(x ) = P(z ) = ________________

c. Find the probability that he makes fewer than 425 free-throws this season. P(r ) = P(x ) = P(z ) = ________________

d. Find the probability that he makes between 425 and 475 free-throws this season. P( ) = P( ) = P( ) = _____________

Solution

Normal Approximation to Binomial Distribution
Mean ( np ) =600 * 0.74 = 444
Standard Deviation ( npq )= 600*0.74*0.26 = 10.7443
Normal Distribution = Z= X- u / sd                   
              
a)
P(X >= 475) = (475-444)/10.7443
= 31/10.7443 = 2.8853
= P ( Z >2.885) From Standard Normal Table
= 0.002
b)
P(X <= 425) = (425-444)/10.7443
= -19/10.7443= -1.7684
= P ( Z <-1.7684) From Standard NOrmal Table
= 0.0385     
c)
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 425) = (425-444)/10.7443
= -19/10.7443 = -1.7684
= P ( Z <-1.7684) From Standard Normal Table
= 0.0385
P(X < 475) = (475-444)/10.7443
= 31/10.7443 = 2.8853
= P ( Z <2.8853) From Standard Normal Table
= 0.99804
P(425 < X < 475) = 0.99804-0.0385 = 0.9595