Calculating Mean and Variance of a Normal Distribution

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SUMMARY

The discussion focuses on calculating the mean (V) and variance (C^2) of a normally distributed random variable X, given specific probabilities. The provided probabilities are P(x>102)=0.42 and P(x<97)=0.25. The calculated mean is 100.8, but the variance was incorrectly stated as (-5.7)^2, which implies an impossible negative standard deviation. The correct interpretation of variance must be a non-negative value.

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jinx007
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Please try this question and see whether you got my answer...i am having some doubt

The random variable X is normally distributed with mean V and variance C^2. It is known that P(x>102)=0.42 and P(x<97)=0.25

calculate V(mean) and variance c^2

i got mean 100.8 and variance(-5.7)^2
 
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You need to show some work, but one comment: if you say the variance is (-5.7)^2, you are implicitly saying the standard deviation is -5.7, which is impossible.
 

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