Question

Find the following z values for the standard normal variable Z. (You may find it useful to reference the z table. Negative values should be indicated by a minus sign. Round your answers to 3 decimal places.) a PIZS z) = 0.1176 b. P(Z SZ50) = 0.1579 c. P(Z > z) = 0.9764 d. P(0.39 SZ sz) = 0.3253

Answer 1

standard Normal a) we get 0.3421 there z is variate and we know that PEze 2] = $(2) and P[azze b]= f(b) d (9) Here p(262) = 0 11176 using standard normal sable P(Z < -1.187 0.1176 on comparing these two we st 17=.=0:187 a) Here P[ZS2Lo] = 6:1579 -> 0)-[2) = 0.15.79 05-5 (Z) = 0.1579 (2) a 2) using standard normal selble we get 0.5 0.1579 - 0°3421 how

on this to above we at Comparing -0.007 23-0 9) P[ z>2] = 0.9764 E-P[2<zj =0.9764 P[Zez Il-0976.420.0236 Now using standard normal table we aut PC 7.5441985) = 0.0236 on companing these two we at hoto Z = -1985 d) A PL 9.39) = 0.3253 » [ 2) - (0.39) 0.3253 27 (2) -0.651730.3253 2) & (2) = 0.6517 +0.3263 -22 2)=0.9.77 how using standard normal table

[11.995) = 0.977 on companity these two wil at Z: 1.999 = 0.1176 23 -1-137 2 = -01407 1)P(252) b). PizçZ40) 20:1579 :c) | P(2>2] = 0.9864 d) P(0.39 $262)=0-3253 7.-1989 21.995