Solving for variable in probability function... HELP Please.

prettylittlepixels

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Begin by finding the value of k that will make the following a probability function:

P(x) = {[(x/2) + 3]^2} / 19k for x = 2, 4, 6, 8, 10

I have several questions to solve using this equation and want to be sure I solve for k properly before making a lot of incorrect calculations on the incorrect value of k.

I'm not sure where to begin with this problem. When I substitute 2 in for the x-value, I end up with 16/19k and if I multiply by k, I get 16/19, which is rounded to 0.84.

Is this right?

Substituting 4 in as the x-value gives me 25/19k, which is completely different. What am I missing?

Thanks in advance?
 
Begin by finding the value of k that will make the following a probability function:

P(x) = {[(x/2) + 3]^2} / 19k for x = 2, 4, 6, 8, 10

I have several questions to solve using this equation and want to be sure I solve for k properly before making a lot of incorrect calculations on the incorrect value of k.

I'm not sure where to begin with this problem. When I substitute 2 in for the x-value, I end up with 16/19k and if I multiply by k, I get 16/19, which is rounded to 0.84.

Is this right?

Substituting 4 in as the x-value gives me 25/19k, which is completely different. What am I missing?

Thanks in advance?

What do you know about what the sum of all the probabilites within a sample space must equal? The sample space here are all the probabilities for those given values of x.
 
What do you know about what the sum of all the probabilites within a sample space must equal? The sample space here are all the probabilities for those given values of x.

It must equal 1. So I need to:

1 = {[(2/2) + 3]^2} / 19k + {[(4/2) + 3]^2} / 19k + {[(6/2) + 3]^2} / 19k + {[(8/2) + 3]^2} / 19k + {[(10/2) + 3]^2} / 19k

1= 16/19k + 25/19k + 36
/19k + 49/19k + 64/19k

1= 190
/19k

Multiply each side of the = by 19k.

19k= 190

Divide each side of the = by 19.

k= 10

Is that right? I'm reviewing for my upcoming final and trying to study the beginning of the semester material I'm a bit rusty at. Thanks for your help!
 
It must equal 1. So I need to:

1 = {[(2/2) + 3]^2} / 19k + {[(4/2) + 3]^2} / 19k + {[(6/2) + 3]^2} / 19k + {[(8/2) + 3]^2} / 19k + {[(10/2) + 3]^2} / 19k

1= 16/19k + 25/19k + 36
/19k + 49/19k + 64/19k

1= 190
/19k

Multiply each side of the = by 19k.

19k= 190

Divide each side of the = by 19.

k= 10

Is that right? I'm reviewing for my upcoming final and trying to study the beginning of the semester material I'm a bit rusty at. Thanks for your help!

Bingo!!
 
Thanks!

So the mean would be 0.2 and the standard deviation is 0.0898= 0.090 (rounded to 3 decimal places)?
 
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Thanks!

So the mean would be 0.2 and the standard deviation is 0.0898= 0.090 (rounded to 3 decimal places)?
Since the smallest possible x is 2 and the largest is 10, the mean has to fall somewhere between.

\(\displaystyle \displaystyle \mathrm E[x] = \sum x\ P(x) = 2\ P(2) + 4\ P(4) + 6\ P(6) + 8\ P(8) + 10\ P(10)\)

Likewise, the expected value of x^2 is

\(\displaystyle \displaystyle \mathrm E[x^2] = \sum x^2\ P(x) = 4\ P(2) + 16\ P(4) + 36\ P(6) + 64\ P(8) + 100\ P(10)\)

The mean and standard deviation are . . .
 
Since the smallest possible x is 2 and the largest is 10, the mean has to fall somewhere between.

\(\displaystyle \displaystyle \mathrm E[x] = \sum x\ P(x) = 2\ P(2) + 4\ P(4) + 6\ P(6) + 8\ P(8) + 10\ P(10)\)

Likewise, the expected value of x^2 is

\(\displaystyle \displaystyle \mathrm E[x^2] = \sum x^2\ P(x) = 4\ P(2) + 16\ P(4) + 36\ P(6) + 64\ P(8) + 100\ P(10)\)

The mean and standard deviation are . . .

Mean= 6
Standard Deviation= 3.162
 
Mean= 6
Standard Deviation= 3.162

I submitted my study guide with all of my work to my professor to prepare for my final next week and he said this part is wrong? He won't give the answer, but just says if it is right or wrong for me to fix. I thought I calculated the mean and standard deviation correctly. I got the value of k=10 correct, but not the mean and standard deviation?
 
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what I got

View attachment 3493

avg is E[x]
v is E[x2]
var is E[x2] - E[x]2

Thanks! I just came on here to say I figured this out!

I plugged x into the original equation and k in as well and got 7.263157895 for the mean and 6.69916897 for the variance because I took the total of 59.45263158 and subtracted the mean squared. Then I took the square root of the variance to get 2.588275289 as the standard deviation.

THANK YOU!
 
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