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MAC 15First, consider an 'experiment' where we will toss a fair die and let  X = the number of dots facing up. of course the possible values for X are 1, 2, 3, 4, 5 and 6 and each of these is equally likely. Show the probability distribution for X using the first two columns in the table below. Avoid rounding errors, show the probabilities as fractions - not as decimals. Next fill in the next two columns using fractions. Don't reduce fractions as you'll want to have a common denominator in each column. * means multiplyx P(X = x)x * P(X = x)x^2 * P(X = x)1   2   3   4   5   6   SUM    a) Express the Expected Value of X, also known as the mean, u, as both a reduced fraction and as a decimal b) show your work to find variance and standard deviation in both fraction and decimal.

Question

MAC 15

First, consider an 'experiment' where we will toss a fair die and let  X = the number of dots facing up. of course the possible values for X are 1, 2, 3, 4, 5 and 6 and each of these is equally likely. Show the probability distribution for X using the first two columns in the table below. Avoid rounding errors, show the probabilities as fractions - not as decimals. Next fill in the next two columns using fractions. Don't reduce fractions as you'll want to have a common denominator in each column. 

* means multiply

P(X = x) x * P(X = x) x^2 * P(X = x)
1      
2      
3      
4      
5      
6      
SUM      

 

a) Express the Expected Value of X, also known as the mean, u, as both a reduced fraction and as a decimal

 

b) show your work to find variance and standard deviation in both fraction and decimal. 

 

check_circleAnswer
Step 1

Consider the random variable X, that is defined as the number of dots facing up in a fair die. The values of X are 1,2,3,4,5 and 6, which are equally likely.

 

The probability of getting X is equal to 1:

Number of favorable cases
P(X Total number of exhaustive cases
1
6
Thus, for rest of the events:
1
1
P(X2)P3)P(X=4)=
P(X= 5) =승,P(X =6) =
P(X 3)
P(X4)
6
6
6
1
help_outline

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Number of favorable cases P(X Total number of exhaustive cases 1 6 Thus, for rest of the events: 1 1 P(X2)P3)P(X=4)= P(X= 5) =승,P(X =6) = P(X 3) P(X4) 6 6 6 1

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Step 2

The completed table is shown below:

xPx)
P(x)
xP(x)
1
2
3
1
4
6
-|0N6mo 0n|0|0| 0
help_outline

Image Transcriptionclose

xPx) P(x) xP(x) 1 2 3 1 4 6 -|0N6mo 0n|0|0| 0

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Step 3

a)

The expected value of X can b...

6
E(X) -ΣxP(x)
1
=1- +2- +3-+4- +5
6
1
1
1
1
6
6
1
6
6
6
21
_
6
7
3.5
7
and decimal form is 3.5
The mean in reduced fraction form is
help_outline

Image Transcriptionclose

6 E(X) -ΣxP(x) 1 =1- +2- +3-+4- +5 6 1 1 1 1 6 6 1 6 6 6 21 _ 6 7 3.5 7 and decimal form is 3.5 The mean in reduced fraction form is

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