Question

What is the probability that Z is between negative 1.53 −1.53 and 1.84 1.84​? The probability that Z is between negative 1.53 −1.53 and 1.84 1.84 is nothing . ​(Round to four decimal places as​ needed.) b. What is the probability that Z is less than negative 1.53 −1.53 or greater than 1.84 1.84​? The probability that Z is less than negative 1.53 −1.53 or greater than 1.84 1.84 is nothing . ​(Round to four decimal places as​ needed.) c. What is the value of Z if only 2 2​% of all possible Z values are​ larger? The value of Z if only 2 2​% of all possible Z values are larger is nothing . ​(Round to two decimal places as​ needed.) d. Between what two values of Z​ (symmetrically distributed around the​ mean) will 94.26 94.26​% of all possible Z values be​ contained? The two values of Z for which 94.26 94.26​% of all possible Z values are contained between are nothing and nothing . ​(Use ascending order. Round to two decimal places as​ needed.)

Answer 1

Solution,

Using standard normal table,

a) P( -1.53 < Z < 1.84)

= P( Z < 1.84) - P( Z < -1.53)

= 0.9671 - 0.0630

= 0.9041

b) P( -1.53 > Z OR Z > 1.84)

= P( Z < - 53 ) + P( Z > 1.84)

= 0.0630 + 1 - P( Z < 1.84)

= 0.0630 + 1 - 0.9671

= 0.0630 + 0.0329

= 0.0959

c) Using standard normal table,

P(Z > z) = 2%

= 1 - P(Z < z) = 0.02  

= P(Z < z) = 1 - 0.02

= P(Z < z ) = 0.98

= P(Z < 2.05 ) = 0.98  

z = 2.05

d) Using standard normal table,

P( -z < Z < z) = 94.26%

= P(Z < z) - P(Z <-z ) = 0.9426

= 2P(Z < z) - 1 = 0.9426

= 2P(Z < z) = 1 + 0.9426

= P(Z < z) = 1.9426 / 2

= P(Z < z) = 0.9713

= P(Z < 1.90) = 0.9713

= z  ± 1.90

z = -1.90, 1.90