Suppose X is a normally distributed random variable with mu

Suppose X is a normally distributed random variable with mu = 50 and sigma = 5 Find a value of the random variable, call it X0. such that p(X LE x0)= .8315 p(40 LE X LE x0)= .8962 25% of the values of X arc less than x0 Need All Work/steps

Solution

a)

First, we get the z score from the given left tailed area. As          
          
Left tailed area =    0.8315      
          
Then, using table or technology,          
          
z =    0.960109714      
          
As x = u + z * s,          
          
where          
          
u = mean =    50      
z = the critical z score =    0.960109714      
s = standard deviation =    5      
          
Then          
          
x = critical value =    54.80054857   [ANSWER]

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b)

First, we get the z score of the left endpoint. As          
          
x1 = left endpoint =    40      
u = mean =    50      
s = standard deviation =    5      
          
Thus,          
          
z1 = (x1 - u) / s = leftt endpoint z score =    -2      
          
Thus, by table/technology, the left tailed area of the left endpoint is          
          
P(z<z2) =    0.022750132      
          
Thus, the left tailed area of the right endpoint is given by          
P(z1<z<z2) =    0.8962      
P(z<z2) = P(z<z1) + P(z1<z<z2) =    0.918950132      
          
Using table or technology, we see that          
z2 = z score of rightt endpoint =    1.398044395      
Thus,          
          
x2 = u + z2*s =    56.99022197   [ANSWER]  

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C)

First, we get the z score from the given left tailed area. As          
          
Left tailed area =    0.25      
          
Then, using table or technology,          
          
z =    -0.67448975      
          
As x = u + z * s,          
          
where          
          
u = mean =    50      
z = the critical z score =    -0.67448975      
s = standard deviation =    5      
          
Then          
          
x = critical value =    46.62755125   [ANSWER]