4h+7=4h-3 i think i have it but im not sure 100% so i need hlp
2 answers:
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Nice job with part A on getting those correct answers.
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Part B
- mu = mean = 43
- sigma = standard deviation = 3
Let's calculate the z score for the raw score x = 37
z = (x - mu)/sigma
z = (37 - 43)/3
z = -6/3
z = -2
A negative z score tells us we're below the mean. Specifically, we are exactly 2 standard deviations under the mean.
If you repeat those steps for x = 46 (not changing mu or sigma), then you should get z = 1. So we're now 1 standard deviation above the mean.
Ultimately, we want to know the area under the standard normal curve (aka z curve) from z = -2 to z = 1.
Refer to the chart below which is a breakdown of the Empirical Rule. The pink areas are within 1 standard deviation of the mean. We have two regions taking up roughly 34% of the full area under the curve, so they combine to 34+34 = 68%
Then we add on another 13.5% which is the blue area on the left (between z = -2 and z = -1)
So 13.5+68 = 81.5% of the area under the curve is between z = -2 and z = 1.
<h3>Answer: 81.5</h3>Note: the percent sign is already taken care of for us, so there's no need to type it in. But the answer above of course means 81.5% and that answer is approximate.
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Part C
Let's find the z score for x = 34. The values of mu and sigma are the same as before.
z = (x - mu)/sigma
z = (34-43)/3
z = -9/3
z = -3
So computing P(X > 34) is the same as P(Z > -3)
Notice how adding up all of the values in the chart mentioned gets us:
2.35+13.5+34+34+13.5+2.35 = 99.7
which means 99.7% of the distribution is within 3 standard deviations of the mean. The remaining 100% - 99.7% = 0.3% is the combined area of both tails. So each tail is (0.3%)/2 = 0.15%
In short, 0.15% of the area under the curve is to the left of z = -3
Therefore, 100% - 0.15% = 99.85% of the widgets are going to have z values larger than -3, which in turn means they have values larger than 34.
<h3>Answer: 99.85</h3>Again you don't have to worry about the percent sign. Like before, the answer is approximate.
