Normal and uniform probability distribution.

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SUMMARY

The discussion focuses on calculating probabilities related to the number of raisins in a cereal box, assuming both normal and uniform distributions. For a normal distribution with a mean of 80 and a standard deviation of 6, the probability of having fewer than 70 raisins is 4.80%, while the probability of having more than 90 raisins is also 4.80%. In contrast, under a uniform distribution, the probability of having fewer than 70 raisins is 87.5%, and the probability of having more than 90 raisins is 0%. The discussion also touches on determining the parameters for a uniform distribution given a mean of 80 and variance of 36.

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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
 

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