Cookies are made in a factory with an average mass of 22 grams and a standard deviation of 4 grams. some cookies are packed in a box. According to the manufacturer, the average mass of a cookie from at least 99% of the boxes is at least 20 grams. How many cookies should the manufacturer at least put in a box?
The answer has to be 22
pls help
thank u
To find the minimum number of cookies the manufacturer should put in a box, we need to determine the number of standard deviations above the mean that corresponds to the desired minimum average mass of 20 grams at a 99% confidence level.
First, let's calculate the z-score using the formula:
z = (x - μ) / σ
where:
x = desired minimum average mass (20 grams)
μ = average mass of cookies from the factory (22 grams)
σ = standard deviation of the cookie masses (4 grams)
z = (20 - 22) / 4
z = -2 / 4
z = -0.5
Next, we need to find the z-score corresponding to the 99th percentile. We can look this up on a standard normal distribution table or use a calculator. The z-score for the 99th percentile is approximately 2.33.
Now we can determine the number of standard deviations between the desired average mass and the mean using:
z = (x - μ) / σ
Rearranging the formula to solve for x, we get:
x = (z * σ) + μ
x = (2.33 * 4) + 22
x ≈ 31.32
So, rounding up to the nearest whole number, the manufacturer should put at least 32 cookies in a box to ensure that at least 99% of the boxes have an average mass of 20 grams or more.
Therefore, the answer should be 32 cookies, not 22.