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Studentka2010 [4]
3 years ago
7

Problem 3 in your textbook presents three different sample spaces for a horse race with five horses running. Let the horses be d

esignated by the letters A, B, C, D, and E. A more complete characterization of an outcome of the horse race would be to designate which horse finishes in first through fifth places
(a) Consider the sample space for the set of outcomes characterized in this way. How many such outcomes are in the sample space?
b) How many outcomes are in the event that horse A finishes first?
(c) If G is the event that horse A finishes first and H is the event that horse B does not finish second, describe in words the event GnH. How many outcomes are in this event?
Mathematics
1 answer:
KiRa [710]3 years ago
5 0

Answer:

a) 120

b) 24

c) 18

Step-by-step explanation:

part a

The sample space is defined by the rank of the horses.

Hence, 5 ranks would permute to = 5! = 120 outcomes

part b

Fixing the first position for horse A we are left with four horses and 4 positions, the position are permutated to 4! = 24 outcomes

part c

Fixing the first position for horse A we are left with four horses and 4 positions, and horse B can not finish second hence:

A _ B _ _  the rest can permute hence, 3! = 6 outcomes

A _ _ B _ the rest can permute hence, 3! = 6 outcomes

A _ _ _ B the rest can permute hence, 3! = 6 outcomes

Total outcomes is sum of 3 cases above = 18 outcomes

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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
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anastassius [24]

Answer:

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Step-by-step explanation:

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Step-by-step explanation:

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faltersainse [42]

3/4=6/8

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Solnce55 [7]

Answer:

The solution of the inequality is x ≤ -4. A graph of the solution should have a vertical line passing through x = -4 and be shaded to the left of x = -4

Step-by-step explanation:

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Subtract 13 from both sides

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Dividing both sides by the -7 changes the inequality sign and we have

x ≤ -4

Hence, the solution of the inequality is x ≤ -4 and the graph of the solution should have a vertical line passing through x = -4 and it should be shaded to the left of x = -4 indicating that only numbers less than or equal to -4 are possible solutions of the inequality.

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6 0
3 years ago
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