A) The answer is 38%.
The event has only two possible outcomes: the top mark is earned either by a man or by a woman.
The events are also independent of each other: the top mark earner does not depend on who earned the second highest mark.
Therefore, we are talking about a binomial distribution, in which the probability of success (mark earned by a woman) is p(W) = 0.38, which means 38%.
B) The probability that exactly 6 of the 10 top marks were earned by women is 9.34%.
The probability of getting exactly 6 women in 10 marks is given by the formula:
[tex]P(X) = \frac{n!}{k!(n-k)!} p^{k}(1 - p)^{n-k} [/tex]
where:
n = total number of events = 10
k = number of success we want = 6
p = probability of a succesfull event = 0.38
Substituting the numbers:
[tex]P(W=6) = \frac{10!}{6!(10-6)!} 0.38^{6}(1 - 0.38)^{10 - 6} \\
= \frac{10!}{6!(4)!} 0.38^{6}(0.62)^{4}[/tex]
P(W = 6) = 210 · 0.00301 · 0.14776
= 0.0934
Hence, the probability of having 6 women earning among the top 10 marks is 0.0934, which means 9.34%
C) We would expect to have 3 women in the top 10.
We would expect that the percentage of the total population is the same of the top 10 marks, therefore:
W = n · p
= 10 · 0.38
= 3.8
Since we cannot have decimals of a physical person, the closest integer is 3.
