Question

You’re a researcher looking at whether or not applicants are accepted to a prestigious law internship. Only 10% of applicants receive a call to a first interview. You’re interested in two samples: all students from a rural community college who applied (24) and all students from Arizona who applied over the last five years (5804).

What are the mean and standard deviation for the large sample?

Explain in the context of this scenario what the mean represents, with appropriate rounding and units.

For the large sample, what is the probability that at least 600 students in the past five years received a first call?

Answer 1

BIONOMIAL DISTRIBUTION
pmf of B.D is = f ( k ) = ( n k ) p^k * ( 1- p) ^ n-k
where   
k = number of successes in trials
n = is the number of independent trials
p = probability of success on each trial
I.
mean = np
where
n = total number of repetitions experiment is executed
p = success probability
mean = 5804 * 0.1
= 580.4
II.
variance = npq
where
n = total number of repetitions experiment is executed
p = success probability
q = failure probability
variance = 5804 * 0.1 * 0.9
= 522.36
III.
standard deviation = sqrt( variance ) = sqrt(522.36)
=22.8552
mean =580.4
standard deviation =22.8552

the probability that at least 600 students in the past five years received a first call
P( X < 600) = P(X=599) + P(X=598) + P(X=597) + P(X=596) + P(X=595) + P(X=594) + P(X=593) + P(X=592) + P(X=591) + P(X=590)    
= ( 5804 599 ) * 0.1^599 * ( 1- 0.1 ) ^5205 + ( 5804 598 ) * 0.1^598 * ( 1- 0.1 ) ^5206 + ( 5804 597 ) * 0.1^597 * ( 1- 0.1 ) ^5207 + ( 5804 596 ) * 0.1^596 * ( 1- 0.1 ) ^5208 + ( 5804 595 ) * 0.1^595 * ( 1- 0.1 ) ^5209 + ( 5804 594 ) * 0.1^594 * ( 1- 0.1 ) ^5210 + ( 5804 593 ) * 0.1^593 * ( 1- 0.1 ) ^5211 + ( 5804 592 ) * 0.1^592 * ( 1- 0.1 ) ^5212 + ( 5804 591 ) * 0.1^591 * ( 1- 0.1 ) ^5213 + ( 5804 590 ) * 0.1^590 * ( 1- 0.1 ) ^5214
= 0.7988
P( X > = 600 ) = 1 - P( X < 600) = 0.2012