Data Questions(Listed in Description)

Question

1.A team of 7 students is to be selected from a class of 30.

a. Aaron, Brad and Charles are students in this class. What is the probability that team includes these 3 students?

b. How many team would not include all three students?

*Would you use indirect reasoning for b?

2.The results for a university entrance exam are normally distributed with µ = 48 and σ = 9 . What is the probability that an exam chosen at random will have a score greater than 63, but less than 73? (Assume data is continuous)

3.By using binomial distribution and normal approximation to answer a question, do you get the same answer from both methods?

Answer 1

Vinu

1. You can choose 7 students from 30 in 30 C 7 ways = 2035800 ways (30 ! / 23 ! 7 !)

(a) If A,B,C are in the team 3 places are taken. You can choose the other team members in 27 C 4 ways = 17550 ways. So probability = 17550 / 2035800 = 0.0086

(b) Well 17550 ways include all three. So how many will not use all three (use the 2035800 and 17550).

1. Well you need p (63<X<73) = p ( (63-48)/9 < Z < ((73-48)/9) = p (1.67 < Z < 2.78). So use a table of areas under the standard normal - find the area when Z = 2.78 and take away the area when Z = 1.67

1. For (3) you will get approximately the same answer. Binomial distribution gives exact answer, normal approximation gives a good approximation (there will be a small error).

Mike

1.