Betty’s Bite-Size Candies are packaged in bags. The number of candies per bag is normally distributed, with a mean of 50 candies and a standard deviation of 3. At a quality control checkpoint, a sample of bags is checked, and 4 bags contain fewer than 47 candies. How many bags were probably taken as samples?
To solve this problem, we need to use the normal distribution formula to find the probability of getting 4 bags with fewer than 47 candies out of a sample of bags.
First, we need to calculate the z-score for 47 candies using the formula:
z = (x - μ) / σ
where x is the actual number of candies (47), μ is the mean number of candies (50), and σ is the standard deviation (3).
z = (47 - 50) / 3 = -1
Next, we need to find the probability of getting a z-score of -1 or lower using a standard normal distribution table or calculator. The probability is approximately 0.1587.
Since we want to find the probability of getting 4 bags with fewer than 47 candies, we need to calculate the probability of getting this result in a sample of bags using the binomial distribution formula:
P(X = 4) = (n choose 4) * p^4 * (1-p)^(n-4)
where n is the number of bags in the sample, p is the probability of getting a bag with fewer than 47 candies (0.1587), and (n choose 4) is the number of ways to choose 4 bags out of n.
We want to find the value of n that makes this probability equal to the observed result (4 bags with fewer than 47 candies):
P(X = 4) = (n choose 4) * 0.1587^4 * (1-0.1587)^(n-4) = 0.04
Using a calculator or trial and error, we find that n = 45 is the closest integer value that gives us a probability of 0.04 or close to it. Therefore, we can conclude that about 45 bags were probably taken as samples.