Question

Let Z be a normal random variable with mean µ = 0 and standard deviation σ = 1, that is, Z ∼ N(0, 1). Find each of the following:

(a) P(Z ≤ −1.13).

(b) P(Z ≥ −2.18).

(c) P(2.13 ≤ Z ≤ 2.57).

(d) P(−2.3 ≤ Z ≤ −1.1).

(e) P(0 ≤ Z ≤ 1.54).

(f) P(−1.54 ≤ Z ≤ 1.54).

(g) N(1.1243).

(h) N(−1.1243).

Answer 1

Answers

All probabilities are obtained using Standard Normal Tables.

(a) P(Z ≤ −1.13) = 0.1292 Answer

(b) P(Z ≥ −2.18) = 0.9854 Answer

(c) P(2.13 ≤ Z ≤ 2.57)

     = P(Z ≤ 2.57) - P(Z ≤ 2.13)

     = 0.9949 – 0.9834

     = 0.0115 Answer

(d) P(−2.3 ≤ Z ≤ −1.1)

     = P(Z ≤ - 1.1) - P(Z ≤ - 2.3)

     = 0.1357 – 0.0107

     = 0.1250 Answer

(e) P(0 ≤ Z ≤ 1.54)

     = P(Z ≤ 1.54) - P(Z ≤ 0)

     = 0.9382 – 0.5

     = 0.4382 Answer

(f) P(−1.54 ≤ Z ≤ 1.54)

    = P(Z ≤ 1.54) - P(Z ≤ - 1.54)

     = 0.9382 – 0.0618

     = 0.8764 Answer

[NOTE: By symmetry property of Normal distribution, P(−1.54 ≤ Z ≤ 0) = P(0 ≤ Z ≤ 1.54). So,   

P(−1.54 ≤ Z ≤ 1.54) = 2 x P(0 ≤ Z ≤ 1.54) = 2 x Answer of (e)]

(g) N(1.1243)

     =P(Z ≤ 1.1243)

     = 0.8931 Answer

(h) N(−1.1243).

     =P(Z ≤ - 1.1243)

     = P(Z ≥ 1.1243) [By symmetry property of Normal distribution]

     = 1 – Answer of (g)

     = 0.1069 Answer

DONE