Question

Suppose a university announced that it admitted 2450 students for the following years freshman class. However, the university has dorm room spots for only 1772 freshman students. If there is a 65% chance that an admitted student will decide to accept the offer and attend this university, what is the approximate probability that the university will not have enough dormitory room spots for the freshman class?

Answer 1

SOLUTION:

Normal Distribution
Mean ( np ) =2450 * 0.65 = 1592.5
Standard Deviation ( √npq )= √2450*0.65*0.35 =23.6088
Normal Distribution = Z= X- u / sd   

P(X < 1772) = (1772-1592.5)/23.6088
= 179.5/23.6088= 7.603
= P ( Z <7.603) From Standard Normal Table
= 0.97725
              
P(X >= 1786) = 1 - 0.97725 = 0.02275.

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