I hope this helps you
Area=6^2pi=36pi
Circumference =2.6pi=12pi
If scores on an exam follow an approximately normal distribution with a mean of 76.4 and a standard deviation of 6.1 points, then the minimum score you would need to be in the top 2% is equal to 88.929.
A problem of this type in mathematics can be characterized as a normal distribution problem. We can use the z-score to solve it by using the formula;
Z = x - μ / σ
In this formula the standard score is represented by Z, the observed value is represented by x, the mean is represented by μ, and the standard deviation is represented by σ.
The p-value can be used to determine the z-score with the help of a standard table.
As we have to find the minimum score to be in the top 2%, p-value = 0.02
The z-score that is found to correspond with this p-value of 0.02 in the standard table is 2.054
Therefore,
2.054 = x - 76.4 ÷ 6.1
2.054 × 6.1 = x - 76.4
12.529 = x - 76.4
12.529 + 76.4 = x
x = 88.929
Hence 88.929 is calculated to be the lowest score required to be in the top 2%.
To learn more about normal distribution, click here:
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Me too maybe b I just need some answers
3/10 times 2/3 3 times 2 is 5 and 10 times 3 is 30 5/30 a
For a 95% confidence interval, the corresponding z-score is 1.96. Therefore the deviation will by 1.96*0.5 lbs = 0.98 lbs. Therefore, the confidence interval will be (5 - 0.98, 5 + 0.98), which is (4.02, 5.98). The weight range is from 4.02 lbs to 5.98 lbs.
