Assume that Z has a standard normal distribution. Determine the value for z that solves each of the following. a. P(-z < Z < z) = 0.95 (Round your answer to two decimal places (e. g. 98.76))
b. P(-z < Z < z) = 0.99 (Round your answer to two decimal places (e. g. 98.76))
c. P(-z < Z < z) = 0.68 (Round your answer to three decimal places (e. g. 98.765))
d. P(-z < Z < z) = 0.9973 (Round your answer to two decimal places (e. g. 98.76))

a)  P  ( - 1.96  < Z < 1.96 )

b) P  ( - 2.58 < Z < 2.58)

c) P (  -0.995  < Z < 0.995 )

d) P  ( - z  <  Z  < z )     =  P ( ( Z ± 3σ )   then that is close to 1

Step-by-step explanation:

a)  P  ( - z  <  Z  < z )     =   P  ( - 1.96  < Z < 1.96 )

CI = 95 %   significance level  α  = 5 %   α  = 0.05    α/2 = 0.025

z = 1.96

b)  P  ( - z  <  Z  < z )     =   P  ( - 2.58 < Z < 2.58)

CI = 99 %   significance level  α  = 1 %   α  = 0.01    α/2 = 0.005

z = 2.58

c) P  ( - z  <  Z  < z )     =   P (  -0.995  < Z < 0.995 )

CI = 68 %   significance level  α  = 32 %   α  = 0.32    α/2 = 0.16

z ≈ 0.9954

We interpolate in this case

1             ⇒  0.1587

0.99      ⇒  0.1611

0.01    ⇒   0.0024        

 x       ⇒    0.0013       x  =  0.01 *0.0013  /  0.0024  

x =  0.005416

and    z =  0.99 + 0.005416

z = 0.9954

d) P  ( - z  <  Z  < z )     =   P (  - 0.00 < Z < 0. 00)

CI = 0.9973 %   significance level  α  = 0.0027 %   α  = 0.000027             α/2 = 0.0000135

z = 0.00003375     ⇒ z = 0.00

NOTE: The value of α  is too small. The Empirical Rule establishes that 99.7 % of all values in a normal distribution fall in the interval ( Z ± 3σ)

that means all the values. Then the probability of finding the random variable between that range is close to 1 and we can not find in tables a number to approximate just with only two decimal places

answer: i think the answer is etheir c or d but i don't know which one but have a good day!

step-by-step explanation:

207/8 is the right