hockeychick23
  • hockeychick23
Find the expected value of the number of questions you'd get right by guessing. What are the variance and standard deviation?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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hockeychick23
  • hockeychick23
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hockeychick23
  • hockeychick23
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hockeychick23
  • hockeychick23
@jim_thompson5910 can you help me please?

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jim_thompson5910
  • jim_thompson5910
X = number of correct questions |dw:1433396801483:dw|
jim_thompson5910
  • jim_thompson5910
multiply X and P(X) to form the X*P(X) column of values
hockeychick23
  • hockeychick23
jim_thompson5910
  • jim_thompson5910
good
jim_thompson5910
  • jim_thompson5910
now add up the values in the X*P(X) column and you will get the expected value
hockeychick23
  • hockeychick23
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jim_thompson5910
  • jim_thompson5910
so the expected value is 0.5 which means, you expect (on average) to get 0.5 questions right if you randomly guessed
hockeychick23
  • hockeychick23
ohhh ok thanks!
jim_thompson5910
  • jim_thompson5910
now onto the second part "What are the variance and standard deviation? "
jim_thompson5910
  • jim_thompson5910
you're going to have 2 additional columns. I'm going to erase the X*P(X) column to make room
jim_thompson5910
  • jim_thompson5910
instead I'm going to have X - mu and (X - mu)^2 |dw:1433397803766:dw|
jim_thompson5910
  • jim_thompson5910
mu = expected value = 0.5
jim_thompson5910
  • jim_thompson5910
oh wait, there's one more column needed the P(X)*(X-mu)^2 column |dw:1433397806736:dw|
hockeychick23
  • hockeychick23
oh ok, so do i find the mean of p(x)?
jim_thompson5910
  • jim_thompson5910
you're asking about the value of mu? mu = 0.5 and it was found when you computed the expected value
hockeychick23
  • hockeychick23
oh ok sorry it took me a while to calculate the numbers
jim_thompson5910
  • jim_thompson5910
this is what I'm getting
jim_thompson5910
  • jim_thompson5910
I see the typo you made. Look at the bottom row, last column
jim_thompson5910
  • jim_thompson5910
you should have (2-0.5) and not (1-0.5)
hockeychick23
  • hockeychick23
sorry i pasted the wrong table this is what i got
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hockeychick23
  • hockeychick23
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jim_thompson5910
  • jim_thompson5910
btw if you have excel or open office (it should work in there too), you can type in formulas like you see attached
jim_thompson5910
  • jim_thompson5910
that way you can automate things a bit
hockeychick23
  • hockeychick23
Ok thanks
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jim_thompson5910
  • jim_thompson5910
now add up everything in the P(X)*(X-mu)^2 column, you should get 0.375
jim_thompson5910
  • jim_thompson5910
0.375 is the variance take the square root of the variance to get the standard deviation
hockeychick23
  • hockeychick23
standard deviation= .61237243569 variance= 0.375
jim_thompson5910
  • jim_thompson5910
looks good
hockeychick23
  • hockeychick23
ok thanks so much! i have a question pertaining to the same data that i need help with if you can help me with it: Let's say that for one of the two questions you can narrow the answer choices down to three. Now your chance of getting that question right by guessing is about .33. What is the new expected value for the number of questions you'd get right by guessing? What are the new variance and standard deviation of this random variable? Drawing another tree diagram and making another probability distribution table may help you answer this question. I'll give you another medal
jim_thompson5910
  • jim_thompson5910
were you able to get started on this one? or no?
hockeychick23
  • hockeychick23
sort of but I'm not sure if i did it right, i did
jim_thompson5910
  • jim_thompson5910
I made question 1 have P(correct) = 0.33 and question 2 has P(correct) = 0.25 |dw:1433401116557:dw|
jim_thompson5910
  • jim_thompson5910
are you seeing how I'm getting these values?
hockeychick23
  • hockeychick23
ohhh yes thanks!
jim_thompson5910
  • jim_thompson5910
so here is getting 0 questions correct |dw:1433401932751:dw| ie getting both wrong
jim_thompson5910
  • jim_thompson5910
0.0825 is cut off, but that's getting both correct |dw:1433401967587:dw|
hockeychick23
  • hockeychick23
Ok so 0.5025 is the probability of getting 0 correct and 0.0825 is getting both correct.
jim_thompson5910
  • jim_thompson5910
and the other values are added to get the probability of getting exactly 1 correct so that gives us this new table |dw:1433402014075:dw|
jim_thompson5910
  • jim_thompson5910
yes
hockeychick23
  • hockeychick23
Ok so would i need to find the expected value again?
hockeychick23
  • hockeychick23
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jim_thompson5910
  • jim_thompson5910
this is what I get
jim_thompson5910
  • jim_thompson5910
You have 0.825 instead of 0.0825
jim_thompson5910
  • jim_thompson5910
So I'm getting an expected value of 0.58 and a variance of 0.4086 standard deviation = sqrt(variance) = sqrt(0.4086) = 0.63921827 (approximate)
hockeychick23
  • hockeychick23
ok now I'm getting that too, thanks! (I'm going to post something and tag you in it to give you another medal)
jim_thompson5910
  • jim_thompson5910
that's fine, it doesn't matter about the medal thing lol
jim_thompson5910
  • jim_thompson5910
I'm glad it's making sense now

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