box: [1 3 5 7]
a estimate the sum of the draws will be more than 1500
b estimate the chance that there will be fewer than ninety 3s –> for this part i changed the box to be [0 3 0 0]
thanks!
1 Answer

a)
If the sum of the draws is greater than 1500, then the average draw must be at least 1500/4 = 3.75
THe actual expected value of the draw is the average of the four numbers, 4.
THis means that the expected sum is 4(400) = 1600.
In order to get a sum of 100 less than this, one would have to be beow the average 100 times and dead on every other time. Since the likelihoods of 1 and 3 are the same and those of 5 ad 7 are the same, if we reduce the problem to “Numbers greater than 4” and “Numbers less than four”.
Since the expected value of numbers less than four is (1+3)/2 = 2 and the expected value of those greater than four is 6, it’s easier to imagine the box as:
[ 2 6 ]
Let x be the number of sixes drawn in 400 draw.
The sum in this case will be 6x + 2(400x) = 800 +4x
Then the the sum is more than 1500 only if
800 + 4x > 1500
4x > 700
x > 175
P(x > 175) = 1 – P(x <= 175) which is (on a TI calculator)
1 – binomcdf(400, 0.5, 175) = 0.9929
b) For this problem we want to call drawing a 3 a “success”, and find the probability of fewer than ninety successes.
The probability of an individual success is 0.25, and we want the cumulative probability of 0 89 successes given 400 draws
Again, using a TI calculator, enter
binomcdf(400, 0.25, 90) = 0.864