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?

Answer 1

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.