I need help with this problem on (Normal Probability Distribution)

Question

Suppose that the Fort Wayne Mad Ants are playing the Cleveland Charge in a basketball game. The number of points scored in a game by the Mad Ants is normally distributed with a mean of 107  and standard deviation of 11 . The number of points scored in a game by the Charge is normally distributed with a mean of  110 and standard deviation of .15

a. What is the probability that the Mad Ants will score more than 100 points in a game (to 2 decimals)?

b. What percentage of games would you expect the Charge to score more than 115  points in a game (to 0 decimal)?

  %

c. What is the probability that the Mad Ants will score more points in a game than the mean number of points scored by the Charge (to 2 decimals)?

Answer 1

Hi Christina,

Assuming points for both teams are normally distributed, this problem involves the classic equation in introductory statistics:

z = (x-mu)/sigma

z=z-score

x=value you are interested in

mu=mean

sigma=standard deviation

a. You are interested in probability of Mad Ants scoring over 100, so:

x= 100

mu= 107

sigma= 11

z= (100-107)/11

z= -0.64

Now. the z-table shows the probability to the left of the curve; that is, the probability that the Mad Ants score less than 100. From z-table:

P(Z<-0.64) = P(X<100) = 0.2611

Now, to get the probability the Mad Ants score greater than 100, we need to apply the Complement Rule aka the "One Minus Trick."

P(X>a) = 1 - P(X<a); a=any real number

Thus:

P(Z>-0.64) = P(X>100) = 1 - 0.2611 = 0.7389;

Rounded to 2 decimals: P = 0.74

b. We set up the same equation for the charge--z= (x-mu)/sigma

x=115

mu= 110

sigma= 15

z= (115-110)/15

z= 0.33

From z-table:

P(Z<0.33)= 0.6293

Again, we use the Complement Rule:

P(Z>0.33) = P(X>=115) = 1 -0.6293 = 0.3707

P = 37%

c. Mean points scored by the charge is 110, so we'll use the same equation with that value in for x, but make sure we use mu and sigma for Fort Wayne, not Cleveland.

x= 110

mu= 107

sigma= 11

z= (110-107)/11

z= 0.27

From z-table:

P(Z<0.27) = 0.6064

Again, Complement Rule:

P(Z>0.27) = P(X>110) = 1-0.6064 = 0.3936;

To two decimals:

P = 0.39

I hope this helps.