Write the number in decimal notation. "Twelve thousand and twelve thousandths"

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

Find the area of the polygon.

svetlana [45]

This is a trapezoid and the trapezoid formula is b1 plus b2/ 2 times height.

17 plus 5 is 22. Divided by 2 is 11.

Now, we have to multiply 11 by the height : 5.

11 times 5 is 55 and that is your area. 55 units squared.

7 0

3 years ago

18 bikes, 9 have gears (G), 11 have suspension (S) and 5 with both,What is the probability that a bike with suspension or gears?

vichka [17]

Answer:

5/9

Step-by-step explanation:

Total number of bikes = 18
Bikes with gears S = 9
Bikes with suspension S = 11
Bikes with both GS = 5

<u>From the information above we have:</u>

Bikes with gears only = 9 - 5 = 4
Bikes with suspension only = 11 - 5 = 6

<u>Sum of the above two gives us the number of bikes with suspension or gears:</u>

4 + 6 = 10

<u>Probability of selected bike be G or S is:</u>

P(G or S) = 10/18 = 5/9

4 0

3 years ago

Find all polar coordinates of point p where p = (6, -pi/5)

Lelu [443]

Given a point in coordinates form $(r,\theta)$ , one can compute the cartesian form like this:
$(r\cos\theta,r\sin\theta )$
We have:
$r=6,\theta=-\dfrac{\pi}{5}$
We get the cartesian form then:
$\left(6\cos-\dfrac{\pi}{5},6\sin-\dfrac{\pi}{5}\right)$

4 0

3 years ago

If 2a - b = 4 , find a.

julia-pushkina [17]

<h2>Answer: a = ¹/₂ (4 + b)</h2>

<h3>Step-by-step explanation:</h3>

To solve for 'a' we have to make it the subject of the equation. Since there are two unknowns ('a' & 'b'), we won't get a numerical value of 'a', but an expression in terms of the second unknown 'b'.

Since 2a - b = 4 <em> [add 'b' to both sides]</em>

then 2a = 4 + b <em>[divide both sides by 2 = multiplying by ¹/₂]</em>

a = ¹/₂ (4 + b)

3 0

3 years ago