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OverLord2011 [107]
3 years ago
15

CAN SOMEONE PLEASE HELP ME WITH THIS STATISTICS PROBLEM I DONT UNDERSTAND IT AT ALLL!!!! The mean of a normal probability distri

bution is 390; the standard deviation is 95. a. Μ ± 1σ of the observations lie between what two values? Lower Value Upper Value b. Μ ± 2σ of the observations lie between what two values? Lower Value Upper Value c. Μ ± 3σ of the observations lie between what two values? Lower Value Upper Value
Mathematics
1 answer:
tankabanditka [31]3 years ago
4 0

Answer:

A.)

Lower Value = 295

Upper Value = 485

B.)

Lower Value = 200

Upper Value = 580

C.)

Lower Value = 105

Upper Value = 675

Step-by-step explanation:

Given that mean (M) = 390

Standard deviation (σ) = 95

Lower Value = mean value - number of standard deviations specified

Upper Value = mean value + number of standard deviations specified

a. Μ ± 1σ of the observations lie between what two values?

Lower Value = 390 - 1(95) = 295

Upper Value = 390 + 1(95) = 485

b. Μ ± 2σ of the observations lie between what two values?

Lower Value = 390 - 2(95) = 200

Upper Value = 390 + 2(95) = 580

c. Μ ± 3σ of the observations lie between what two values?

Lower Value = 390 - 3(95) = 105

Upper Value = 390 + 3(95) = 675

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9514 1404 393

Answer:

  A to D is about 2087 ft

Step-by-step explanation:

For the portion of the path of interest, the sine and cosine relations apply.

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  Cos = Adjacent/Hypotenuse

__

AB is the adjacent side to the angle marked 33°, with hypotenuse 975 ft.

  cos(33°) = AB/(975 ft)

  AB = (975 ft)cos(33°) ≈ 817.70 ft

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CD is the hypotenuse of the right triangle BCD. The side BC that we know is opposite the given angle, so the sine relation applies.

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  CD = BC/sin(46°)

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__

Now we know the segments of the path of interest:

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  A–D = 817.70 ft + 531.02 ft + 738.21 ft = 2086.93 ft

  A–D ≈ 2087 ft

_____

<em>Additional comments</em>

The attachment shows all of the segments computed working counterclockwise from A. The angles in triangle DEF are computed using the Law of Sines from sides DF and DE and angle DEF. Segments EG and GH are computed using the Law of Cosines.

When we get to points F, G, H, we find that there is some inconsistency with the locations that would be computed working clockwise using AB and angle BFA. This inconsistency shows up in an error in the 119° angle in the attached figure.

This means that segments on the back side of the route, along path EFGHA, will vary somewhat depending on how they're computed.

We assume that points B, C, F, G are collinear.

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