Don't guess!

Suppose that the data for analysis includes the attributeage. Theagevalues for the datatuples are (in increasing order) 13, 15,

Bas_tet [7]

Answer:

a) $\bar X = \frac{\sum_{i=1}^{27} X_i }{27}= \frac{809}{27}=29.96$

$Median = 25$

b) $Mode = 25, 35$

Since 25 and 35 are repeated 4 times, so then the distribution would be bimodal.

c) $Midrange = \frac{70+13}{3}=41.5$

d) $Q_1 = \frac{20+21}{2} =20.5$

$Q_3 =\frac{35+35}{2}=35$

e) Min = 13 , Q1 = 20.5, Median=25, Q3= 35, Max = 70

f) Figura attached.

g) When we use a quantile plot is because we want to show the percentage or the fraction of values below or equal to an specified value for the distribution of the data.

By the other hand the quantile-quantile plot shows the quantiles of the distribution values against other selected distribution (specified, for example the normal distribution). If the points are on a straight line we assume that the data values fit very well to the hypothetical distribution selected.

Step-by-step explanation:

For this case w ehave the following dataset given:

13, 15, 16, 16, 19, 20, 20, 21, 22, 22, 25, 25, 25, 25, 30,33, 33, 35, 35, 35, 35, 36, 40, 45, 46, 52, 70.

Part a

The mean is calculated with the following formula:

$\bar X = \frac{\sum_{i=1}^{27} X_i }{27}= \frac{809}{27}=29.96$

The median on this case since we have 27 observations and that represent an even number would be the 14 position in the dataset ordered and we got:

$Median = 25$

Part b

The mode is the most repeated value on the dataset on this case would be:

$Mode = 25, 35$

Since 25 and 35 are repeated 4 times, so then the distribution would be bimodal.

Part c

The midrange is defined as:

$Midrange = \frac{Max+Min}{2}$

And if we replace we got:

$Midrange = \frac{70+13}{3}=41.5$

Part d

For the first quartile we need to work with the first 14 observations

13, 15, 16, 16, 19, 20, 20, 21, 22, 22, 25, 25, 25, 25

And the Q1 would be the average between the position 7 and 8 from these values, and we got:

$Q_1 = \frac{20+21}{2} =20.5$

And for the third quartile Q3 we need to use the last 14 observations:

25, 30,33, 33, 35, 35, 35, 35, 36, 40, 45, 46, 52, 70

And the Q3 would be the average between the position 7 and 8 from these values, and we got:

$Q_3 =\frac{35+35}{2}=35$

Part e

The five number summary for this case are:

Min = 13 , Q1 = 20.5, Median=25, Q3= 35, Max = 70

Part f

For this case we can use the following R code:

> x<-c(13, 15, 16, 16, 19, 20, 20, 21, 22, 22, 25, 25, 25, 25, 30,33, 33, 35, 35, 35, 35, 36, 40, 45, 46, 52, 70)

> boxplot(x,main="boxplot for the Data")

And the result is on the figure attached. We see that the dsitribution seems to be assymetric. Right skewed with the Median<Mean

Part g

When we use a quantile plot is because we want to show the percentage or the fraction of values below or equal to an specified value for the distribution of the data.

By the other hand the quantile-quantile plot shows the quantiles of the distribution values against other selected distribution (specified, for example the normal distribution). If the points are on a straight line we assume that the data values fit very well to the hypothetical distribution selected.

6 0

4 years ago

5. Find the triangle similar to AABC at the right.

Alex Ar [27]

Answer:

the answer is b

Step-by-step explanation:

7 0

3 years ago

I really don't understand help

Zielflug [23.3K]

Answer:

128 ft sq.

Step-by-step explanation:

The way you find the area of a square is by multiplying one side by another, and since it is a square, the sides are the same. This means that if we have a square with a side length of x, the way to find the area of that square is to multiply x by x, which can be written as $x^2$ . Now, since we're doubling this, we can say that double Mary's garden is $2 * x^2$ . If we substitute 8 in for x, we have to multiply $8^2$ by 2 or 8 by 8 by 2 to find our answer. 8 times 8 is 64 and 64 times 2 is 128.