Normal and uniform probability distribution.

Kinetica
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Am I right? Thank you!Problem:
A cereal brand of cereal claims that the mean number of raisins in each box is 80 with a standard deviation of 6. If the raisins are normally distributed, what are the chances that an arbitrary box has

1) fewer than 70 raisins and
2) more than 90 raisins.

What should be your answers if the raisins are uniformly distributed.

Solution:

{70-80}/{6}=-1.67; we disregard the negative sign in order to find the value in the table. The value is 0.9520. 1-0.9520=0.048 or 4.80% chance that an arbitrary box has fewer than 70 raisins.

{90-80}/{6}=1.67, which is also 0.9525.
1-0.9520=0.048 or 4.80% chance that an arbitrary box has more than 90 raisins. If the raisins are uniformly distributed, the probability that an arbitrary box has fewer than 70 raisins is 1/80 * 70=0.875 or 87.5%.
Probability that there are more than 90 raisins is 0..
 
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Kinetica said:
Am I right? Thank you!


Problem:
A cereal brand of cereal claims that the mean number of raisins in each box is 80 with a standard deviation of 6. If the raisins are normally distributed, what are the chances that an arbitrary box has

1) fewer than 70 raisins and
2) more than 90 raisins.

What should be your answers if the raisins are uniformly distributed.

Solution:

{70-80}/{6}=-1.67; we disregard the negative sign in order to find the value in the table. The value is 0.9520. 1-0.9520=0.048 or 4.80% chance that an arbitrary box has fewer than 70 raisins.

{90-80}/{6}=1.67, which is also 0.9525.
1-0.9520=0.048 or 4.80% chance that an arbitrary box has more than 90 raisins.


If the raisins are uniformly distributed, the probability that an arbitrary box has fewer than 70 raisins is 1/80 * 70=0.875 or 87.5%.
Probability that there are more than 90 raisins is 0..

You need to find the uniform distribution with mean 80 and variance 62= 36. Do you know the formulas for mean and variance of a uniform distribution on the interval [a,b]? Once you have determined a and b the rest is straightforward.

RGV
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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