in a particular hospital, 60% of the staff is male and 40% female. In a random sample of 50 staff members, what is the probability that more than half will be female?

Answer 1

To solve this problem, we can use the binomial probability formula.

In a random sample of 50 staff members, the probability that more than half will be female can be calculated by finding the probability of having 26, 27, 28, 29, ..., 50 female staff members and summing up these individual probabilities.

Let's denote the probability of selecting a female staff member as p.

p = 0.40 (since 40% of the staff is female)
q = 1 - p = 0.60 (since 60% of the staff is male)

The formula for the probability of obtaining k successes in n trials is:

P(X = k) = (n C k) * p^k * q^(n-k)

Where (n C k) is the combination or binomial coefficient, n! / (k! * (n - k)!)

Therefore, to calculate the probability of more than half being female, we need to calculate the sum of probabilities for k = 26 to 50.

P(X > 25) = P(X = 26) + P(X = 27) + P(X = 28) + ... + P(X = 50)

P(X > 25) = Σ (n C k) * p^k * q^(n-k) for k = 26 to 50

Calculating this sum would be time-consuming. However, we can use complementary probability to simplify the calculation.

The complementary probability approach is: P(X > 25) = 1 - P(X ≤ 25)

P(X ≤ 25) = P(X = 0) + P(X = 1) + P(X = 2) + ... + P(X = 25)

P(X ≤ 25) = Σ (n C k) * p^k * q^(n-k) for k = 0 to 25

Now, let's calculate P(X ≤ 25):

P(X ≤ 25) = Σ (50 C k) * (0.40)^k * (0.60)^(50-k) for k = 0 to 25

Using a calculator or statistical software, we can calculate this sum as approximately 0.912.

Therefore, P(X > 25) = 1 - P(X ≤ 25) = 1 - 0.912 = 0.088

Hence, the probability that more than half of the random sample of 50 staff members will be female is approximately 0.088 or 8.8%.