The least weight of a bag in the top 5 percent of the distribution is; 246
From the complete question below, we are given;
Population mean; μ = 240
Population standard deviation; σ = 3
Z-score formula is;
z = (x' - μ)/σ
- Now, we want to find the least weight in the top 5 of the distribution and as such we will use;
1 - 0.05/2 = 0.025 as significance level
Z-score at significance level of 0.025 is 1.96
Thus;
1.96 = (x' - 240)/3
3 × 1.96 = x' - 240
x' = 240 + 5.88
x' = 245.88
Approximating to a whole number gives;
x' = 246
Complete question is;
A machine is used to fill bags with a popular brand of trail mix. The machine is calibrated so the distribution of the weights of the bags of trail mix is normal, with mean 240 grams and standard deviation 3 grams. Of the following, which is the least weight of a bag in the top 5 percent of the distribution?
Read more about z-score at; brainly.com/question/25638875
Answer:
See explanation below.
Explanation:
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The correct answer is C
We are solving for C so we need to isolate C
Multiply 2 to both sides
2W = CV^2
Divide V^2
C = 2W/V^2
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Answer:
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