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. P(Z S z) 0.1020 b. P(z s Zs0) 0.1772 c. P(Z> z)-0.9929 d. P(0.40 sZz)0.3368 1.270 2.450

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Answer #1

Solution:

Given that,

a ) P( Z \leq z ) = 0.1020

Using standard normal table

z = - 1.270

b )   P(z \leq Z  \leq 0 ) = 0.1772

P ( Z  \leq 0 ) - P ( Z  \leq z  ) = 0.1772

Using standard normal table

P ( Z  \leq z  ) = 0.5000 +  0.1772

P ( Z  \leq z  ) = 0.6772

z = 0.460

c )  P ( Z >  z  ) = 0.9929

= 1 - P ( Z < z  ) = 0.9929

Using standard normal table

P ( Z < z  ) = 1 -  0.9929

P ( Z < z  ) = 0.0071

z = - 2.452

d )   P( 0.40  \leq Z  \leq z ) = 0.3368

P ( Z  \leq z ) - P ( Z  \leq 0.40 ) = 0.3368

Using standard normal table

P ( Z  \leq z  ) = 0.6554 + 0.3368

P ( Z  \leq z  ) = 0.9922

z = 2.418

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