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Bond [772]
3 years ago
7

Consider a sample with data values of 27, 25, 20, 15, 30, 34, 28, and 25. Compute the range, interquartile range, variance, and

standard deviation (to a maximum of 2 decimals, if decimals are necessary).
Mathematics
1 answer:
Valentin [98]3 years ago
6 0

So we have a set of integers,

X=\{15,20,25,25,27,28,30,34\}

And we need to compute some stats.

First the range,

The range of data set is difference between the maximum and the minimum of the data set.

So the minimum is 15 and maximum is 34 therefore the range is 34 - 15 = 19.

Second the interquartile range,

The interquartile range of data set is the difference of the first and third quartiles.

First quartile is the value separating the lower quarter and higher three - quarters of the set. The first quartile is computed by taking median of the lower half of the sorted set. This is 15, 20, 25, 25 and it's median is 22.5

Third quartile is therefore the median of 27, 28, 30, 34 which is 29.

Next up, variance.

The variance of set measures how much data is spread out. For data set x_1,\dots, x_n with an average a,

\mathsf{var}(X)=\Sigma_{i=1}^{n}\dfrac{(x_i-a)^2}{n-1}

So first compute the average value which is actually a mean,

a=\dfrac{1}{2}\Sigma_{i=1}^{n}a_i

The sum \Sigma_{i=1}^{n}a_i of numbers in set is 204.

Divide this by number of elements in the set (8).

a=\dfrac{204}{8}=25.5

Then compute the variance,

\mathsf{var}(X)=\dfrac{\Sigma_{i=1}^{n}(x_i-a)^2}{n-1}\approx\boxed{34.57}

And finally standard deviation,

Since have computed variance the standard deviation is nothing too hard. It's defined as a square root of variance,

\mathsf{sde}(X)=\sqrt{\dfrac{\Sigma_{i=1}^{n}(x_i-a)^2}{n-1}}=\sqrt{34.57}\approx\boxed{5.88}

Hope this helps.

r3t40

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iris [78.8K]

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Step-by-step explanation:

By definition, an Isosceles triangle has two equal sides and its opposite angles are congruent.

Observe the figure attached, where the isosceles triangle is divided into two equal right triangles.

So, in this case you need to use the following Trigonometric Identity:

sin\alpha =\frac{opposite}{hypotenuse}

In this case, you can identify that:

\alpha =\angle BAD=\angle CAD=18\°\\\\opposite=BD=CD=x\\\\hypotenuse=AB=AC=24

Substituting values, and solving for "x", you get:

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Therefore, the length of BC rounded to nearest tenth, is:

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prisoha [69]

Answer:

y²=4√2.x

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The distance between the focus and the directrix of the parabola P is

\frac{ |0-4| }{\sqrt{(1)^{2}+(-1)^{2}  } }=\frac{4}{\sqrt{2} }

{Since the perpendicular distance of a point (x1, y1) from the straight line ax+by+c =0 is given by \frac{ |ax1+by1+c| }{\sqrt{a^{2}+b^{2}  } } }

Let us assume that the equation of the parabola which is congruent with parabola P is y²=4ax

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⇒ a=√2

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3 years ago
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Answer:

Thanks, but here is a problem I'll give myself.

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2. Answer: X > 1

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