Find x such that PX x 09382 Round z value to 2 decimal plac

Find x such that P(X x) = 0.9382. (Round \"z\" value to 2 decimal places, and final answer to nearest whole number.)

Find x such that P(X > x) = 0.025. (Round \"z\" value to 2 decimal places, and final answer to nearest whole number.)

Find x such that P(3,900 X x) = 0.1217. (Round \"z\" value to 2 decimal places, and final answer to nearest whole number.)

Find x such that P(X x) = 0.484. (Round \"z\" value to 2 decimal places, and final answer to nearest whole number.)

Solution

a)
P ( Z < x ) = 0.9382
Value of z to the cumulative probability of 0.9382 from normal table is 1.54
P( x-u/s.d < x - 3900/2400 ) = 0.9382
That is, ( x - 3900/2400 ) = 1.54
--> x = 1.54 * 2400 + 3900 = 7596                  
b)
P ( Z > x ) = 0.025
Value of z to the cumulative probability of 0.025 from normal table is 1.96
P( x-u/ (s.d) > x - 3900/2400) = 0.025
That is, ( x - 3900/2400) = 1.96
--> x = 1.96 * 2400+3900 = 8604                  
d)
P ( Z < x ) = 0.484
Value of z to the cumulative probability of 0.484 from normal table is -0.04
P( x-u/s.d < x - 3900/2400 ) = 0.484
That is, ( x - 3900/2400 ) = -0.04
--> x = -0.04 * 2400 + 3900 = 3804