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standard deviation and probabilities?
100 numbers are rounded off to the nearest integer and then summed. Suppose the individual round-off error are uniformly distributed over (-.5, .5).
Remember a random variable that follows a uniform distribution over (a,b) has a mean of (a+b)/2, and variance of (1/12)(b-a)^2.
What is the mean of the round-off error of one number? 0 is the answer
What is the standard deviation of the round-off error of one number? I need help with this one.
Suppose the sum of the round-off errors is Y.
What is the mean of Y? 0 is the answer
What is the standard deviation of Y? I need help with this one.
To compute the probability that the resultant sum of the 100 numbers differs from the exact sum by more than 3, we find the two tails on the distribution of Y.
What is the z-score at Y=3 ? . 3/sqrt(100/12) is the answer
What is the probability that the rounded sum is larger than the exact sum by more than 3 ? I need help with this one.
What is the probability that the rounded sum differs from the exact sum by more than 3 ? I need help with this one.
2 Answers
- cidyahLv 77 years ago
Variance = (b-a)^2 / 12 = (0.5 - (-0.5) )^2 / 12 = 1/12
standard deviation of one number = 1/sqrt(12)
Variance of y = 100 (1/12) = 100/12 = 25/3
standard deviation of y = 5/sqrt(3)
- ?Lv 67 years ago
the variance of Y=100*(b-a)^2/12=100*1/12=100/12
standard deviation= square root variance
The desired probability=1-Pr[-3<Y<3]=
1-Pr[-3/sqrt(100/12)<Z<3/sqrt(100/12)] since the mean of Y is 0
=1-Pr[-1.04<Z<1.04]=1-(0.85083005-0.14916995)=
1- 0.701660099= 0.298339901
