# A normal random variable x has an unknown mean and standard deviation. the probabilty that x exceeds 4 is 0.9699 Find mean and standard deviation

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We are given that x is a normal random variable from a set with an unknown mean and standard deviation. We are also given that the probability that x exceeds 4 is 0.9699.

This problem is not well defined.

If the probability that a number exceeds 4 is 0.9699, this equates to a standard `z` score of -1.88. We know that `z=(x-bar(x))/s` .

Solving for `s` we get `s=-4/1.88+bar(x)/1.88` . We need for `bar(x)/1.88>4 ==> bar(x)>7.52`

Ex: Let `bar(x)=8` ; then `s=-4/1.88+8/1.88=2.13`

Then `z=(4-8)/2.13=-1.88` and thus the probability of x>4 is the area to the right of z=-1.88 under the standard normal curve, or 0.9699

Ex: Let `bar(x)=16` ; then `s=-4/1.88+16/1.88=6.38`

Then `z=(4-16)/6.38=-1.88` etc...

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**The mean can be any number greater than 7.52. Then the standard deviation will be `s=-4/1.88+bar(x)/1.88` **

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