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?
A. 25 bags
B. 12 bags
C. 8 bags
D. 5 bags
To solve this problem, we first need to calculate the z-score for the number of candies in each bag that contained fewer than 47 candies:
z = (X - μ) / σ
z = (47 - 50) / 3
z = -1
Next, we can look up the probability of a z-score of -1 on a standard normal distribution table, which is approximately 0.1587. This means that the probability of a bag containing fewer than 47 candies is 0.1587.
Since 4 bags were found to contain fewer than 47 candies in the sample, we can use the binomial distribution formula to calculate the probability of this happening:
P(x) = nCx * p^x * (1-p)^(n-x)
Where:
n = number of bags sampled
x = number of bags with fewer than 47 candies
p = probability of a bag containing fewer than 47 candies
Given:
x = 4,
p = 0.1587
Now we need to find the number of bags sampled:
P(4) = nC4 * 0.1587^4 * (1-0.1587)^(n-4)
Since we don't have a specific number of bags provided, let's try plugging in the options given to see which one fits:
A. 25 bags:
nC4 * 0.1587^4 * (1-0.1587)^(25-4) = 0.008
This is not the correct answer.
B. 12 bags:
nC4 * 0.1587^4 * (1-0.1587)^(12-4) = 0.177
This is not the correct answer.
C. 8 bags:
nC4 * 0.1587^4 * (1-0.1587)^(8-4) = 0.365
This is not the correct answer.
D. 5 bags:
nC4 * 0.1587^4 * (1-0.1587)^(5-4) = 0.302
This is the closest approximate answer.
Therefore, the answer is D. 5 bags were probably taken as samples.