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Lina20 [59]
3 years ago
9

Anyone can help me with this

Mathematics
1 answer:
Leokris [45]3 years ago
8 0

Given:

The data values are

11, 12, 10, 7, 9, 18

To find:

The median, lowest value, greatest value, lower quartile, upper quartile, interquartile range.

Solution:

We have,

11, 12, 10, 7, 9, 18

Arrange the data values in ascending order.

7, 9, 10, 11, 12, 18

Divide the data in two equal parts.

(7, 9, 10), (11, 12, 18)

Divide each parenthesis in 2 equal parts.

(7), 9, (10), (11), 12, (18)

Now,

Median = \dfrac{10+11}{2}

            = \dfrac{21}{2}

            = 10.5

Lowest value = 7

Greatest value = 18

Lower quartile = 9

Upper quartile = 12

Interquartile range (IQR) = Upper quartile - Lower quartile

                                        = 12 - 9

                                        = 3

Therefore, median is 10.5, lowest value is 7, greatest value is 18, lower quartile 9, upper quartile 12 and interquartile range is 3.

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Answer:

See explanation below.

Step-by-step explanation:

Numerical way

We can calculate the mean and the standard deviation from a sample with the following formulas:

\bar X = \frac{\sum_{i=1}^n X_i}{n}

s= \frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}

We have the following dataset for women: 34,34,26,37,42,41,35,31,41,33,30,74,33,49,38,61,21,41,26,80,72

And the mean and deviation are:

\bar X_F = 41.857

s_F = 16.399

We can calculate the median ordering the data like this:

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And the median would be on the 11 position of the data ordered, for this case is:

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And the range is given by: Range_F = 80-21=59

We can also calculate the coefficient of variation given by:

\hat{CV} =\frac{s}{\bar X} = \frac{16.399}{41.857}=0.392

We have the following dataset for men:

43, 40,48,48,56,38,60,32,40,42,37,76,39,55,45,35,61,33,51,32,42

And the mean and deviation are:

\bar X_M = 45.381

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And the range is given by: Range_M = 76-32=44

We can also calculate the coefficient of variation given by:

\hat{CV} =\frac{s}{\bar X} = \frac{11.218}{45.381}=0.247

We se that the mean for the male groups is higher than the mean for female group. The deviation is higher for the female group comapred to the male group. We have more variation for the female group as we can see on the range and the coeffcient of variation.

Graph

We can see the histograms for each group on the figures attached.

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