Question

a. Find z such that 24% of the standard normal curve lies to the right of z.

b. Find the z values such that 64% of the standard normal curve lies between –z and z

c. Find z such that 36% of the standard normal curve lies to the left of z

Answer #1

solution

Using standard normal table,

P(Z > z) = 24%

= 1 - P(Z < z) = 0.24

= P(Z < z) = 1 - 0.24

= P(Z < z ) = 0.76

= P(Z < 0.71 ) = 0.76

z =0.71 ( using z table )

(B)

middle 64% of score is

P(-z < Z < z) = 0.64

P(Z < z) - P(Z < -z) = 0.64

2 P(Z < z) - 1 = 0.64

2 P(Z < z) = 1 + 0.64 = 1.64

P(Z < z) =1.64 / 2 = 0.82

P(Z < 0.92) = 0.82

z ± 0.92 using z table

(C)

Using standard normal table,

P(Z < z) = 36%

= P(Z < z) = 0.36

= P(Z <-0.36 ) = 0.36

z =- 0.36 Using standard normal z table,

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