Question

5.78

A selective college would like to have an entering class of 1000 students. Because not all studets who are offered admission accept, the college admits more than 1000. Past experience shows that about 83% of the students admitted will accept. The college decides to admit 1200 students.Assuming that students make their decisions independently, the number who accept his the B(1200, 0.83) distribution. If this number is less than 1000, the college will admit students from its waiting list.

a What are the mean and standard deviation of the number X of students who accept?

b. Use the Normal approximation to find the probability that at least 800 students accept.

c. The college does not want more than 1000 students. What is the probability that more than 1000 will accept?

d. If the college decides to decrease the number of admission offers to 1150, what is the probability that more than 1000 will accept?

Answer 1

Solution:

a) n = 1200
p = 0.83
For this case, applying normal approximation to binomial, we get:
mean = n*p= 1200 *0.83 = 996
variance = n*p*(1-p) = 1200 * 0.83 * 0.17 = 169.32
std dev = sqrt(variance) = sqrt(169.32) = 13.0123

b) Find : P( X >= 800) = P( X >799.5) (this is continuity correction and is required if X only takes integer values)
We convert to standard normal form, Z ~ N(0,1)
by z1 = (x1 - u )/s

z1 =( x - u)/s = (799.5 - 996)/13.0123 = -15.1

P( X > x1) = 1 - P( X < x1) = 1 - P( Z < z1)
= 1 - P( Z < -15.1)
= 1 _ _ _ _ _ _ _ _ _ (From Z-table)
c)

Find : P( X >= 1000) = P( X >999.5) (this is continuity correction and is required if X only takes integer values)
We convert to standard normal form, Z ~ N(0,1)
by z1 = (x1 - u )/s

z1 =( x - u)/s = (999.5 - 996)/13.0123 = 0.3

P( X > x1) = 1 - P( X < x1) = 1 - P( Z < z1)
= 1 - P( Z < 0.3)
= 0.3821 _ _ _ _ _ _ _ _ _ (From Z-table)

d) mean=1150*0.83=954.5

standard deviation=sqrt(1150*0.83*(1-0.83))=12.7383
Find : P( X >= 1000) = P( X >999.5) (this is continuity correction and is required if X only takes integer values)
We convert to standard normal form, Z ~ N(0,1)
by z1 = (x1 - u )/s

z1 =( x - u)/s = (999.5 - 954.5 )/12.7383 = 3.53

P( X > x1) = 1 - P( X < x1) = 1 - P( Z < z1)
= 1 - P( Z < 3.53)
= 0.0002 _ _ _ _ _ _ _ _ _ (From Z-table)