Question

Let Z have the Standard Normal distribution.

(a) Find z so that the area under the density curve and to the left of z is .4266.

(b) Find z so that the area under the density curve and to the right of z is .8051.

(c) Find the z–scores of the bounds for the middle 37% of area.

(d) Find z so that P(Z < z) = .7416.

(e) Find z so that P(−z < Z < z) = .4573.

(f) Find z so that P(−1.54 < Z < z) = .7895.

Answer 1

Solution

Using standard normal table,

(a)

P(Z < z) = 0.4266

P(Z < -0.185) = 0.4266

z = -0.185

(b)

P(Z > z) = 0.8051

1 - P(Z < z) = 0.8051

P(Z < z) = 1 - 0.8051 = 0.1949

P(Z < -0.86) = 0.1949

z = -0.86

(c)

P(Z < z) = 0.37

P(Z < -0.33) = 0.37

Z -scores are : -0.33 and +0.33

(d)

P(Z < z) = 0.7416

P(Z < 0.65) = 0.7413

z = 0.65

(e)

P(−z < Z < z) = 0.4573

P(-z < Z < z) = 0.4576
P(Z < z) - P(Z < z) = 0.4573
2P(Z < z) - 1 = 0.4573
2P(Z < z) = 1 + 0.4573
2P(Z < z) = 1.4573
P(Z < z) = 1.4573 / 2 = 0.72865
P(Z < 0.61) = 0.72865
z = 0.61

(f)

P(−1.54 < Z < z) = 0.7895

P(Z < z) - P(Z < -1.54) = 0.7895

P(Z < z) =  P(Z < -1.54) + 0.7895 = 0.0618 + 0.7895 = 0.8513

P(Z < z) = 0.8513

P(Z < 1.04) = 0.8513

z = 1.04