Question

If P ( − 1.4 < z < b ) = 0.7352 , find b.

Please show solution steps and how to enter on a TI-83. Thank you!

Answer 1

P(-1.4<z<b) = 0.7352

P(-1.4<Z<b) = P(b) - P(-1.4)

P(-1.4) = 0.0808 use normcdf function to calculate this from TI-83 calculator

press 2nd then VARS now press 2 then enter normcdf(5,1.4) which will give you the result -0.080756 which means the area is on the negative side. so P(-1.4) = 0.0808.

P(b) -0.0808 = 0.7352

P(b) = 0.8160

from the normal table we get b = 0.90

b = 0.90

calculate upto P(b) = 0.8160 by using TI-83 to calculate b do the following process

press 2nd then VARS now press 3 then enter invNorm(0.816,0,1) yow will get a result 0.9002

b = 0.90