Question

4.17. for a standard normal random variable z, compute (a) p(z ≥ 0.99) (b) p(z ≤ −0.99) (c) p(z < 0.99) (d) p(|z| > 0.99) (e) p(z < 10.0) (f) p(z > 10.0) (g) with probability 0.9, variable z is less than what?

Answer 1

Answer: 1.28

Explanation:

1. P(Z > 0.99) = 1 -Ф (0.99) = 1-0.8389 = 0.1611

2.P(Z <-0.99) = Ф(-0.99) = 0.161 (known that from [a] by symmetry)

3.P(Z <0.99) = Ф(0.99) = 0.8389

4. P(Z >0.99) = P(Z <-0.99) + P(Z > 0.99) = 2(0.1611) = 0.3222

The value z = 10.0 is out because it is too large. On comparing with the largest value, z = 3.99,

we get

5. P(Z < 10.0) > P(Z <3.99) = 1.0

6.P(Z > 10.0)<P(Z > 3.99) =10.0

7. On solving the equation

Ф(Z) = 0.9

for Z.

In Table A4, find the value of z corresponding to the probability 0.9. The nearest value is Ф(1.28) = 0.8997. Therefore, z ≈ 1.28