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...
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%.
To find the probability that fewer than 475 toasters need to be returned, we can use the binomial distribution.
In this case, we have a batch of 25,000 toasters, and the manufacturer estimates a 2% chance that one toaster will need to be sent back for repairs. This means the probability of a toaster not needing repairs is 1 - 0.02 = 0.98.
Let's define our variables:
n = 25,000 (number of trials/toasters)
p = 0.02 (probability of success/not needing repairs)
x = number of toasters that need to be returned (we want fewer than 475)
Now, we want to find the probability P(X < 475), which represents the cumulative probability up to 474 (P(X <= 474)).
To calculate this probability, we can use a binomial probability formula or a statistical calculator. I'll explain both methods:
Method 1: Binomial Probability Formula
The binomial probability formula is:
P(X = x) = (n choose x) * p^x * (1 - p)^(n - x)
To find P(X < 475), we need to calculate the cumulative sum of probabilities for all x values less than 475. It can be a bit time-consuming to calculate and add up each individual probability, so let's use a calculator or statistical software.
Method 2: Statistical Calculator or Software
You can use a calculator or statistical software (e.g., Excel, R, Python) to calculate binomial probabilities. These tools have built-in functions or libraries that provide convenient ways to calculate probabilities.
For example, in Excel, you can use the function BINOM.DIST to calculate the cumulative probability:
=BINOM.DIST(474, 25000, 0.02, TRUE)
This formula calculates P(X <= 474) for given n, p, and x values. The TRUE argument specifies that we want a cumulative probability.
Using these methods, you can find the probability that fewer than 475 toasters need to be returned in a batch of 25,000 toasters, given the estimated 2% return rate.
To solve this problem, we can use the binomial probability formula.
The formula for the binomial probability is:
P(x) = (nCx) * p^x * (1-p)^(n-x)
Where:
- P(x) is the probability of getting exactly x successes
- nCx is the number of combinations of n items taken x at a time
- p is the probability of success for one trial
- (1-p) is the probability of failure for one trial
- n is the total number of trials
In this case, we want to find the probability of fewer than 475 toasters needing to be returned, so we need to calculate the sum of probabilities for x = 0 to 474.
Using the given information:
n = 25,000 toasters
p = 0.02 (2% chance that one toaster needs to be sent back)
Let's calculate the probability using this information: