Question

Assume that 40% of graduates come from University A and and earn salaries of 30,000 on average with a standard deviation of 5,000. The remainder come from University B and earn salaries of 25,000 on average with a standard deviation of 7,500. If you are told that a graduate is earning less than 35,000, what is the probability that they came from University A

Answer 1

38.18% probability that they came from University A

Explanation:

To solve this question, we need to understand the normal probability distribution and the conditional probability formula.

Normal probability distribution:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the zscore of a measure X is given by:

`Z = (X - \mu)/(\sigma)`

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Conditional probability formula:

We use the conditional probability formula to solve this question. It is

`P(B|A) = (P(A \cap B))/(P(A))`

In which

P(B|A) is the probability of event B happening, given that A happened.

`P(A \cap B)` is the probability of both A and B happening.

P(A) is the probability of A happening.

If you are told that a graduate is earning less than 35,000, what is the probability that they came from University A

So:

Event A: earning less than 35,000

Event B: coming form university A.

Probability of earning less than 35,000.

40% come from University A.

University A: Mean 30,000 and standard deviation 5,000.

This probability is the pvalue of Z when X = 35,000. So

`Z = (X - \mu)/(\sigma)`

`Z = (35,000 - 30,000)/(5,000)`

`Z = 1`

`Z = 1` has a pvalue of 0.8413.

Intersection is coming from university A and earning less than 35,000. So

`P(A \cap B) = 0.40*0.8413 = 0.33652`

60% come from university B:

Mean 25,000 and standard deviation 7,5000.

`Z = (X - \mu)/(\sigma)`

`Z = (35,000 - 25,000)/(7,500)`

`Z = 1.33`

`Z = 1.33` has a pvalue of 0.9082.

Then

`P(A) = 0.40*0.8413 + 0.6*0.9082 = 0.88144`

Finally:

`P(B|A) = (P(A \cap B))/(P(A)) = (0.33652)/(0.88144) = 0.3818`

38.18% probability that they came from University A