The random variable X is normally distributed with mean `mu` and standard deviation `sigma` . It is given that P(X > 58.37) = 0.02 and P(X < 40.85) = 0.01, the mean and standard deviation have to be determined.

P(X > 58.37) = 0.02 and P(X < 40.85) = 0.01

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The random variable X is normally distributed with mean `mu` and standard deviation `sigma` . It is given that P(X > 58.37) = 0.02 and P(X < 40.85) = 0.01, the mean and standard deviation have to be determined.

P(X > 58.37) = 0.02 and P(X < 40.85) = 0.01

=> P( X < 58.37) = 0.98 and P(X < 40.85) = 0.01

The z-score corresponding to a probability of 0.98 is 2.05 and the z-score corresponding to a probability of 0.01 is -2.33

This gives `2.05 = ( 58.37 - mu)/sigma`

=> `sigma = (58.37 - mu)/2.05`

Substitute in `-2.33 = (40.85 - mu)/sigma`

=> `-2.33 = ((40.85 - mu)*2.05)/(58.37 - mu)`

=> `-2.33(58.37 - mu) = (40.85 - mu)*2.05`

=> `-2.33*58.37 + 2.33*mu = 40.85*2.05 - 2.05*mu`

=> `mu(2.33 + 2.05) = 2.33*58.37 + 40.85*2.05`

=> `mu = 50.179`

`sigma = 3.995`

**The mean is 50.179 and the standard deviation is 3.995**