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Distribution Probability. Put a lot of thought into it..
- Thread starter dhs316
- Start date
I have a few questions I need help with.
A1. X is N(mu; sigma). Determine mu and sigma so that P(X <= 10) = .60 and P(X =<15) = .70.
A2. X is N(mu; sigma). Determine mu and sigma so that P(X <= 10) = .20 and P(X >=12) = .2.
A3. X is N(17; 4). It is desired to find as short as possible an interval [a; b] such that P(a <= X <=b) =.90. What should a and b be?
A1) For A1, I found two Z score equations [(X-mu)/sigma], plugged in the respective x values, set the equations equal to each other, and solved for the two unknown variables. Would this be the correct method to find the answer?
A2)I was wondering if the z-score equation for P(X>=12) would be [(12+mu)/sigma] , noting that instead of the subtracting, I’m adding since the X is more than or equal to 12. From here, I simply solved in a similar manner as I did in the last problem.
I was also thinking if I could use the fact that P(X<12)=1-P(X>=12) = 1-.2 = .8. With this, I could set the second z score equation equal to .8 with X=12.
Both ways give me different answers, so I am not sure if either method is correct. Can you let me know which train of thought is correct?
A3) For this problem, I used the equation found on the bottom of page 128. Knowing that the whole equation is equal to .9, I plugged in the mu(17) and sigma(2) values into the equation and found that b-a=1.8, which I interpreted as that any values two values that are 1.8 apart will satisfy the equation. However, I am confused if this is actually the shortest interval that the question statement specifies.
Also, I was thinking it may be possible to solve this by looking at the normal distribution tables, but again, I am not sure how the shortest interval would be found. Is there a specific equation that finds shortest intervals? Please let me know if either of my thoughts are correct.
Your approach to Problem 1 is correct.
For both Problems 2 and 3, it is useful to study the normal distribution curve. This may allow for some insights/shortcuts into finding solutions.
Note that in A2, we have a symmetry situation: there are areas of “.2” in both the right and left hand tails of the distribution. Since it is a normal distribution (not skewed), we know right away that the mean is equidistant from both the specified values, 10 and 12. Therefore, the mean is 11.
For A3, it must be recognized that most of the area under the curve lies near the mean. If we go one standard deviation to either side of the mean (two standard deviations apart) and examine the area between these values, we enclose about 68% of the area under the curve. However, if we examine the area between the mean and two standard deviations below the mean, we only enclose about 47.5% of the area under the curve. In both cases the two points chosen were two std deviations apart, but the area they enclose is not the same. From these observations we can conclude that the shortest interval [a, b] will occur when “a” and “b” are symmetrically placed on either side of the mean. Make sense?