This problem involves the binomial distribution, which can be approximated by the normal distribution when n is sufficiently large and the success probability p is not too close to 0 or 1. In this case, n = 125 and p = 0.065, which satisfies these conditions. We can use the normal approximation to find the probabilities requested.
The mean of the binomial distribution is μ = np = 125 x 0.065 = 8.125, and the standard deviation is σ = sqrt(np(1-p)) = sqrt(125 x 0.065 x 0.935) = 2.343.
A. To find the probability of 80 or fewer successes, we need to calculate the z-score for 80:
z = (80 - 8.125) / 2.343 = 33.932
Since we are looking for the probability of a value less than or equal to 80, we need to find the area to the left of this z-score. Using a table or calculator, we find this probability to be almost 1 (i.e., P(z < 33.932) ≈ 1). Therefore, the probability of 80 or fewer successes is approximately 1.
B. To find the probability of 82 or fewer successes, we need to calculate the z-score for 82:
z = (82 - 8.125) / 2.343 = 34.985
Again, we need to find the area to the left of this z-score. This probability is also almost 1 (i.e., P(z < 34.985) ≈ 1), so the probability of 82 or fewer successes is also approximately 1.
C. To find the probability of 75 or more successes, we can either calculate the probability of 74 or fewer (and subtract it from 1), or we can use the z-score for 75:
z = (75 - 8.125) / 2.343 = 29.664
The area to the left of this z-score is almost 1 as well (i.e., P(z < 29.664) ≈ 1), so the probability of 75 or more successes is approximately 1.
In summary, we can approximate the probabilities as follows:
A. P(80 or fewer) ≈ 1
B. P(82 or fewer) ≈ 1
C. P(75 or more) ≈ 1
A sample of 125 is drawn from population with proportion equal to 0.065.determine the probability of observing :
A, 80 or fewer successes.
B, 82 or fewer successes.
C, 75 or more successes.
1 answer
1 answer