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iris [78.8K]
4 years ago
10

A population is normally distributed with mean 18 and standard deviation 1.7. (a) Find the intervals representing one, two, and

three standard deviations of the mean.
Mathematics
1 answer:
elena55 [62]4 years ago
6 0

Answer:

Interval within 1 standard deviation: (16.3,19.7)

Interval within 2 standard deviation: (14.6, 21.4)

Interval within 3 standard deviation: (12.9, 23.1)

Step-by-step explanation:

We are given the following in the question:

Population mean, \mu = 18

Standard deviation, \sigma = 1.7

We have to find the following intervals:

Interval within 1 standard deviation:

\mu \pm 1\sigma\\=18 \pm 1.7\\=(16.3, 19.7)

Interval within 2 standard deviation:

\mu \pm 2\sigma\\=18 \pm 2(1.7)\\= 18 \pm 3.4\\=(14.6,21.4)

Interval within 3 standard deviation:

\mu \pm 3\sigma\\=18 \pm 2(1.7)\\= 18 \pm 5.1\\=(12.9,23.1)

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Step-by-step explanation:

we have to calculate the volume of 2 pipes, actually.

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and then we need to subtract the volume of the inner, hole part from the general pipe volume.

and that is then the volume of the actual pipe material, which we will then "translate" to weight based on the given ratio of 1/0.05 mm³/gm

so, what is the volume of a pipe ? it is actually a cylinder. it has a circular base area, and the length is, of course, the height.

the area of a circle is pi×r².

and the cylinder volume is base area times height :

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also to remember : 1cm = 10mm

so, length = height = 70×10 = 700mm.

it is always important to use the same dimension of numbers when combining them in a calculation.

Vi (inner volume) = pi×12²×700 mm³ = pi×144×700 mm³

Vt (total volume) = pi×13²×700 mm³ = pi×169×700 mm³

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