Question
A manufacturer ships toasters in cartons of 50. In each carton, they estimate a 2% chance that one of the toasters will need to be sent back for minor repairs. In a batch of 25,000 toasters, what are the chances that fewer than 475 need to be returned?
The probability that fewer than 475 toasters need to be returned is...
The probability that fewer than 475 toasters need to be returned is...
Answers
Bot
To solve this problem, we can use the binomial probability formula:
P(X = k) = (n choose k) * p^k * (1-p)^(n-k)
where n is the number of trials (toasters), k is the number of successes (toasters needing repairs), and p is the probability of success (2% chance of needing repairs).
We want to calculate the probability that X < 475, which means we need to find the sum of the probabilities for k = 0 to k = 474.
First, let's find the expected value (mean) and standard deviation of this binomial distribution:
Expected value (mean) = n * p = 25000 * 0.02 = 500
Standard deviation = sqrt(n * p * (1-p)) = sqrt(25000 * 0.02 * 0.98) ≈ 22.07
Now, we'll use the normal distribution to approximate the binomial distribution. We'll convert X = 474.5 (we add 0.5 for continuity correction) to a z-score:
z = (X - mean) / standard deviation = (474.5 - 500) / 22.07 ≈ -1.156
Now, we can use a standard normal table or calculator to find the probability that z < -1.156, which gives us the probability that fewer than 475 toasters need to be returned.
Using a standard normal table or calculator, we find that P(z < -1.156) ≈ 0.1241.
Therefore, the probability that fewer than 475 toasters need to be returned is approximately 12.41%.
P(X = k) = (n choose k) * p^k * (1-p)^(n-k)
where n is the number of trials (toasters), k is the number of successes (toasters needing repairs), and p is the probability of success (2% chance of needing repairs).
We want to calculate the probability that X < 475, which means we need to find the sum of the probabilities for k = 0 to k = 474.
First, let's find the expected value (mean) and standard deviation of this binomial distribution:
Expected value (mean) = n * p = 25000 * 0.02 = 500
Standard deviation = sqrt(n * p * (1-p)) = sqrt(25000 * 0.02 * 0.98) ≈ 22.07
Now, we'll use the normal distribution to approximate the binomial distribution. We'll convert X = 474.5 (we add 0.5 for continuity correction) to a z-score:
z = (X - mean) / standard deviation = (474.5 - 500) / 22.07 ≈ -1.156
Now, we can use a standard normal table or calculator to find the probability that z < -1.156, which gives us the probability that fewer than 475 toasters need to be returned.
Using a standard normal table or calculator, we find that P(z < -1.156) ≈ 0.1241.
Therefore, the probability that fewer than 475 toasters need to be returned is approximately 12.41%.