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3 years ago

a pine tree that is 140 feet tall is added to the sample. how will this outlier affect the shape of the box plot?

Mathematics

2 answers:

enot [183]3 years ago

4 0

Answer:

D. The new box plot will be positively skewed.

Step-by-step explanation:

We have been given a box plot which represents the the heights (in feet) of a sample of pine trees.

We are asked to determine that affect of adding a pine tree that is 140 feet tall to the sample.

We can see from our box plot that the tallest pine tree is 75 feet and 140 is approximately 2 times greater than 75.

After adding the a tree that is 140 feet tall, the maximum height will be 140 and our right whisker will be at point 140.

Since we know that for a positive skewed data, the tail of the distribution is longer on the right hand side.

As 140 is a extreme positive outlier for our given data set, therefore, after adding a tree that is 140 feet tall our box plot will be positively skewed and option D is the correct choice.

Alex3 years ago

3 0

Answer:

D. The new box plot will be positively skewed.

Step-by-step explanation:

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A random sample of 10 shipments of stick-on labels showed the following order sizes.10,520 56,910 52,454 17,902 25,914 56,607 21

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Confidence Interval: (21596,46428)

Step-by-step explanation:

We are given the following data set:

10520, 56910, 52454, 17902, 25914, 56607, 21861, 25039, 25983, 46929

Formula:

$\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}$

where $x_i$ are data points, $\bar{x}$ is the mean and n is the number of observations.

$Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}$

$Mean =\displaystyle\frac{340119}{10} = 34011.9$

Sum of squares of differences = 551869365.6 + 524322983.6 + 340111052.4 + 259528878 + 65575984.41 + 510538544 + 147644370.8 + 80512934.41 + 64463235.21 + 166851472.4 = 2711418821

$S.D = \sqrt{\frac{2711418821}{9}} = 17357.09$

Confidence interval:

$\mu \pm t_{critical}\frac{\sigma}{\sqrt{n}}$

Putting the values, we get,

$t_{critical}\text{ at degree of freedom 9 and}~\alpha_{0.05} = \pm 2.2621$

$34011.9 \pm 2.2621(\frac{17357.09}{\sqrt{10}} ) = 34011.9 \pm 12416.20 = (21595.7,46428.1) \approx (21596,46428)$

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