SOLUTION: The mean of a random variable X is E(X) = a, and the variance is D(X) = b. Find the probability P(X^2 < c). (a = -0.09, b = 1.22, c = 1.77)

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Question 1201861: The mean of a random variable X is E(X) = a, and the variance is D(X) = b. Find the probability P(X^2 < c).
(a = -0.09, b = 1.22, c = 1.77)

Answer by ikleyn(52781)   (Show Source): You can put this solution on YOUR website!
.
The mean of a random variable X is E(X) = a, and the variance is D(X) = b.
Find the probability P(X^2 < c).
(a = -0.09, b = 1.22, c = 1.77)
~~~~~~~~~~~~~~~~~~~


The probability  P(X^2 < c)  is the same as the probability  P(  < X <  ),

which is  P = normalcdf(, , a, ), if you use a regular calculator

with standard function normalcdf, standing for normal cumulative distribution function.


Using given data, you can write


    P = normalcdf(, , a, }) = normalcdf(, , -0.09, ) = 

      = normalcdf(-1.3304, 1.3304, -0.09, 1.1045) = 0.7701.    ANSWER

Solved.



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