replace_school_principal

Replace School Principal

In a study of educational institutions, it was observed that 60% of schools that significantly improved their standardized test scores over the last three years replaced their principals during that period. Conversely, only 35% of schools that did not show significant improvement in test scores replaced their principals. Given that the probability of a school achieving a significant improvement in test scores is 4%, what is the probability that a school, which decides to replace its principal, will also experience a significant improvement in test scores ?

Solution

Let's solve the problem using the given information.
Define the events as follows:

A: The principal is replaced.
B: The school achieves a significant improvement in standardized test scores (more than 5%).

Given:

\( P(A|B) \): Probability that the principal is replaced given that the school achieves a significant improvement in test scores = 60% = 0.60

\( P(A | \neg B) \): Probability that the principal is replaced given that the school does not achieve a significant improvement in test scores = 35% = 0.35

\( P(B) \): Probability that the school achieves a significant improvement in test scores = 4% = 0.04

We need to find P(B|A), the probability that the school achieves a significant improvement in test scores given that the principal is replaced.
Using Bayes' theorem: \[P(B|A) = \frac{P(A|B) \times P(B)}{P(A)}\]
We first need to find P(A), the probability that the principal is replaced:
\[P(A) = P(A \cap B) + P(A \cap \neg B)\] \[P(A) = P(A|B) \times P(B) + P(A| \neg B) \times P( \neg B)\] \[P(A) = 0.60 \times 0.04 + 0.35 \times (1 - 0.04)\] \[P(A) = 0.024 + 0.35 \times 0.96\] \[P(A) = 0.024 + 0.336\] \[P(A) = 0.36\]
Now, we can find P(B|A):
\[P(B|A) = \frac{0.60 \times 0.04}{0.36}\] \[P(B|A) = \frac{0.024}{0.36}\] \[P(B|A) = 0.0667\] Therefore, the probability that a school, which decides to replace its principal, will also experience a significant improvement in standardized test scores is approximately 6.67%.