Hi Hamza,
a-1. This would simply be 11% of the total number of students taking the test. This is also considered the sample mean, or x̅, pronounced "x bar".
.11 * 75 = x̅
a-2. To get the standard deviation of a proportion you use the formula,
σp = √[ P(1 - P) / n ]
where,
σp = standard deviation of proportion
P = proportion that fail
(1 - P) = proportion that pass
n = size of sample
This gives us the standard deviation in terms of a % of the total sample. To get the standard deviation in terms of students we then have to multiply σp by the sample size
σp * n = σsample
b. For this you could use the formula for combinations and permutations, but they want you to find the z-score at which only 4 students would fail.
since the mean is somewhere around 7.6, roughly, then you would be using a left-one-tailed test to find out the z-score needed so that only 4 students would fail. Then you use that z-score to find the probability of getting any value to the left of that z-score.
c. This is just the probability of getting any value to the right of the previous z-score
Let me know if that helped or if you need any further clarification!
