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
We can use the normal distribution to find the probability of a bag containing fewer than 47 candies:
$P(X<47) = P\left(Z < \frac{47 - 50}{3}\right) = P(Z < -1)$
Using a standard normal table or calculator, we find that $P(Z < -1) \approx 0.1587$.
So the probability that a randomly selected bag contains fewer than 47 candies is about 0.1587.
If we assume that the bags are sampled independently, then the number of bags containing fewer than 47 candies follows a binomial distribution with parameters $n$ (number of bags sampled) and $p=0.1587$ (probability of a bag containing fewer than 47 candies).
We want to find the value of $n$ that makes it likely to observe 4 or more bags containing fewer than 47 candies. In other words, we want to find the smallest value of $n$ such that:
$P(X \ge 4) = 1 - P(X < 4) \approx 0.05$
Using a binomial distribution table or calculator, we can find that $P(X \ge 4) \approx 0.0495$ when $n = 25$. So the answer is A. 25 bags were probably taken as samples.
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To find out how many bags were probably taken as samples, we can use the concept of the normal distribution.
Given that the mean number of candies in a bag is 50 and the standard deviation is 3, we can use the formula for a Z-score to determine the probability of a bag having fewer than 47 candies.
The Z-score formula is:
Z = (X - μ) / σ
Where X is the observed value, μ is the mean, and σ is the standard deviation.
In this case, X = 47, μ = 50, and σ = 3.
So, let's calculate the Z-score:
Z = (47 - 50) / 3
Z = -1
Now, we need to find the probability associated with this Z-score using a Z-table. The probability of a Z-score being less than -1 is the same as the probability of having fewer than 47 candies in a bag.
Looking up this probability in the Z-table, we find that the probability is approximately 0.1587.
Given that 4 bags were found to have fewer than 47 candies, to determine how many bags were probably taken as samples, we divide the number of bags found by the probability.
Number of bags taken as samples = 4 / 0.1587 ≈ 25.22
Since we cannot have a fractional number of bags, we can round up to the nearest whole number. Therefore, the answer is approximately 25 bags.
So, the most likely answer is A. 25 bags.