Which data set has an outlier? 25, 36, 44, 51, 62, 77 3, 3, 3, 7, 9, 9, 10, 14 8, 17, 18, 20, 20, 21, 23, 26, 31, 39 63, 65, 66,
umka21 [38]
It's hard to tell where one set ends and the next starts. I think it's
A. 25, 36, 44, 51, 62, 77
B. 3, 3, 3, 7, 9, 9, 10, 14
C. 8, 17, 18, 20, 20, 21, 23, 26, 31, 39
D. 63, 65, 66, 69, 71, 78, 80, 81, 82, 82
Let's go through them.
A. 25, 36, 44, 51, 62, 77
That looks OK, standard deviation around 20, mean around 50, points with 2 standard deviations of the mean.
B. 3, 3, 3, 7, 9, 9, 10, 14
Average around 7, sigma around 4, within 2 sigma, seems ok.
C. 8, 17, 18, 20, 20, 21, 23, 26, 31, 39
Average around 20, sigma around 8, that 39 is hanging out there past two sigma. Let's reserve judgement and compare to the next one.
D. 63, 65, 66, 69, 71, 78, 80, 81, 82, 82
Average around 74, sigma 8, seems very tight.
I guess we conclude C has the outlier 39. That one doesn't seem like much of an outlier to me; I was looking for a lone point hanging out at five or six sigma.
They give the formula as:
Surface Area =<span> (2 • <span>π <span>• r²) + (2 • <span>π • r • height)</span></span></span></span>
However the 2*PI*r^2 part of the formula is used to calculate the 2 "ends" of a cylinder. Since the problem states that you are NOT to count any of surface area of the "ends" then you only need the <span>(2 • <span>π • r • height) part of the formula.
So, r = 3 inches and height = 8 * 3 inches, the side area equals
2 * PI * r * height
2 * 3.14159 * 3 * 24 =
</span></span>
<span>
<span>
<span>
452.39 cubic inches which is the lateral area.</span></span></span>
