# Four hundred draws will be made at random with replacement from the following box…?

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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!

• 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(400-x) = 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

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