Question

A certain early response pregnancy test claims to be able to detect the pregnancy hormone in 88% of pregnant women, 5 days prior to their expected period. In a sample of 25 pregnant women who took the test 5 days before their period, what is the probability that exactly 88% of them saw a positive test result on their pregnancy test?

Answer 1

The probability that exactly 88% of the 25 pregnant women see a positive test result can be calculated using the binomial probability formula.

Step-by-step explanation:

The probability that exactly 88% of the 25 pregnant women see a positive test result can be calculated using the binomial probability formula. In this case, the probability of success (seeing a positive test result) is 0.88, and the number of trials (pregnant women) is 25. To calculate the probability, we use the formula:

P(X = k) = C(n, k) * p^k * (1 - p)^(n - k)

where:

• P(X = k) is the probability of getting exactly k successes
• C(n, k) is the number of ways to choose k successes from n trials (binomial coefficient)
• p is the probability of success in each trial
• k is the number of successes we're interested in
• n is the number of trials

In this case, we'll calculate:

P(X = 0.88*25) = C(25, 0.88*25) * 0.88^(0.88*25) * (1 - 0.88)^(25 - 0.88*25)

Using a calculator or software, you can find that the probability is approximately 0.1992, or 19.92%.