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1 year ago

7

Suppose you are told that a distribution is said to be approximately normal because outliers have skewed the data. List the poss

ible ways outliers can skew an approximately normal distribution and what effect they can have.

1 answer:

andrew-mc [135]1 year ago

8 0

Answer:

Step-by-step explanation:

Outline are values which are entirely different from those remaining values in a data set. These extreme values can skew an approximately normal distribution by skewing the distribution in the direction of the outliers and this makes it difficult for the data set to be analyzed.

Its effect is such that the mean becomes extremely sensitive to extreme outliers making it possible that the mean is this not a representative of the population and this theoretically affects the standard deviation.

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A trail runner gets a new map of her favorite mountain. • Her old map has a scale of 1 cm to 100 m. • Her new map has a scale of

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1 year ago

The base of an aquarium with given volume V is made of slate and the sides are made of glass. If the slate costs seven times as

lubasha [3.4K]

Answer:

$l = \sqrt[3]{\frac{2V}{7}}$ $b = \sqrt[3]{\frac{2V}{7}}$ $h = \sqrt[3]{\frac{49V}{4}}$

Step-by-step explanation:

Represent the volume of the box with V and the dimensions with l, b and h.

The volume (V) is:

$V = l * b * h$

Make h the subject of the formula

$h = \frac{V}{lb}$

The surface area (S) of the aquarium is:

$S = lb + 2(lh + bh)$

Where lb represents the area of the base (i.e. slate):

The cost (C) of the surface area is:

$C = 7 * lb + 1 * 2(lh + bh)$

$C = 7lb + 2(lh + bh)$

$C = 7lb + 2h(l + b)$

Substitute $\frac{V}{lb}$ for h in the above equation

$C = 7lb + 2*\frac{V}{lb}(l + b)$

$C = 7lb + \frac{2V}{lb}(l + b)$

$C = 7lb + \frac{2V}{b} + \frac{2V}{l}$

Differentiate with respect to l and with respect to b

$C_l=7b - \frac{2V}{l^2}$ $=0$

$C_b=7l - \frac{2V}{b^2}$ $=0$

To solve for b and l, we equate both equations and set l to b (to minimize the cost)

$7b - \frac{2V}{l^2}=7l - \frac{2V}{b^2}$

$7l - \frac{2V}{l^2}=7b - \frac{2V}{b^2}$

By comparison:

$l =b$

$C_l=7b - \frac{2V}{l^2}$ $=0$ becomes

$7l - \frac{2V}{l^2}=0$

$7l = \frac{2V}{l^2}$

Cross Multiply

$7l^3 = 2V$

Solve for l

$l^3 = \frac{2V}{7}$

$l = \sqrt[3]{\frac{2V}{7}}$

Recall that: $l =b$

$b = \sqrt[3]{\frac{2V}{7}}$

Also recall that:

$h = \frac{V}{lb}$

$h = \frac{V}{\sqrt[3]{\frac{2V}{7}}*\sqrt[3]{\frac{2V}{7}}}$

$h = \frac{V}{\sqrt[3]{\frac{4V^2}{49}}}$

Apply law of indices

$h = \sqrt[3]{\frac{49V^3}{4V^2}}$

$h = \sqrt[3]{\frac{49V}{4}}$

The dimension that minimizes the cost of material of the aquarium is:

$l = \sqrt[3]{\frac{2V}{7}}$ $b = \sqrt[3]{\frac{2V}{7}}$ $h = \sqrt[3]{\frac{49V}{4}}$

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1 year ago