SOLUTION: Let X be a continuous random variable with probability density function {{{f(x)=(1/sqrt(2pi))e^(-x^2/2)}}} Find P(-1 < X < 1). [Hint: You may want to use a CAS to evaluate this.]

Algebra ->  Probability-and-statistics -> SOLUTION: Let X be a continuous random variable with probability density function {{{f(x)=(1/sqrt(2pi))e^(-x^2/2)}}} Find P(-1 < X < 1). [Hint: You may want to use a CAS to evaluate this.]      Log On


   

Question 1136607: Let X be a continuous random variable with probability density function f%28x%29=%281%2Fsqrt%282pi%29%29e%5E%28-x%5E2%2F2%29 Find P(-1 < X < 1). [Hint: You may want to use a CAS to evaluate this.]
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!

This function models the normal distribution curve when mu = 0 and sigma = 1, aka this is the standard Z distribution


The general template is

Plug in mu = 0 and sigma = 1 to get this slightly simpler equation



which is exactly what you were given to start with.


Find the area under the curve that is to the left of z = -1.00 using a table such as this one. You should find that P(Z < -1.00) = 0.1587

Using that same table, you should also find P(Z < 1.00) = 0.8413

Subtract the areas to get the region between the proper z values we want
P(-1 < Z < 1) = P(Z < 1) - P(Z < -1)
P(-1 < Z < 1) = 0.8413 - 0.1587
P(-1 < Z < 1) = 0.6826


Answer: 0.6826 (which is approximate)