A small Normal Distribution problem

Hi Mr Koh,

                 Here is a problem from my tutorial which I experienced difficulty solving:

                Given that X~N( -12, 6) , find  k such that P( |X+12| < k) =0.85

Student X

I will offer two slightly different methods for your consideration.

1. The distribution of X+12 will have a different mean value; however its variance remains unchanged.

   ie  X+12~N( 0, 6)

   P( |X+12| < k) =0.85 ====> P (-k < X+12 < k ) =0.85

If you were to sketch the normal distribution curve for X+12, and pencil in the limits -k/+k , you will notice that the area under the curve

between -k and +k is equal to 0.85. Since this curve is symmetrical about the y-axis, you can further reinterpret the probability structure

to arrive at P( X+12 < -k) = (1-0.85)/2 = 0.075.

Thereafter, using the Inverse Normal computation, you should arrive at k=3.5261  (shown)

2.  P( |x+12| < k) =0.85 ====> P (-k < X+12 < k ) =0.85 

or  P (-k-12 < X < k -12 ) =0.85  ----------(1)

     Consider standidization to the standard normal Z variable where Z~ N(0, 1), 

     (1) therefore becomes  (-k/sqrt(6) < Z < k/sqrt(6) ) =0.85

     Since the standard normal distribution curve is symmetrical about the y-axis, then it is also alternatively true to state that

P( Z< -k/sqrt(6) )  = (1-0.85)/2 =0.075

    Thereafter, using the Inverse  Normal computation, you shall obtain

-k/sqrt(6) = -1.4395  ===> k= 3.5261  (shown)

Hope this helps. Peace.

Best Regards,

Mr Koh