Let X be normally distributed with mean 2700 and standard d

Let X be normally distributed with mean = 2,700 and standard deviation = 700.

Find x such that P(2,700 X x) = 0.1217. (Round \"z\" value to 2 decimal places, and final answer to nearest whole number.)

Solution

Answer : x =2917

Mu = 2700

SD = 700

P(2700 < X < x) = 0.1217

= > P(X < x) - P(X < 2700) = 0.1217

Now,

For , P(X<2700)

z = (2700 - 2700)/700 = 0

P(X<2700) = 0.5

Therefore,

P(X < x) - P(X < 2700) = 0.1217

=> P(X < x) - 0.5 = 0.1217

=> P(X < x) = 0.6217

Again, for x

z = (x -2700)/700

=> 0.31 = (x -2700)/700 {From z table at z=0.31 , P = 0.6217}

=> x = (0.31*700) + 2700 = 2917 Answer