Question

3. Using the same table from class (or Appendix C in Priviterra, pp. C1-C4), find the proportion under the standard normal curve that lies between each of the following points:

The mean and z =+2.00
The mean and z = 0
z = -1.96 and z = -1.64
z = -.82 and z = + .82
z = +0.50 and z = +1.90

Answer 1

Solution

Back-up Theory

If Z ~ N(0, 1) i.e., Z has Standard Normal Distribution,

Mean of Z = 0 ………………………………………………………………………………………….. (1)

Standard deviation of Z = 1…………………………………………………………………………….. (2)

P(Z < - t) = P(Z > t) ……………………………………………………………………………………. (3)

And in general, for any variable, X, P(t1 < X < t2) = P(X < t2) - P(X < t1) …………………………. .(4)    

Now to work out the solution,

Answers are given directly in the following table. Detailed working follows the table.

(a)	The mean and z =+2.00	0.472
(b)	The mean and z = 0	0
(c)	z = -1.96 and z = -1.64	0.0255
(d)	z = -.82 and z = + .82	0.4878
(e)	z = +0.50 and z = +1.90	0.2798

Detailed working

Note : All probabilities are read off from the Standard Normal Table giving P(Z > t) for t ≥ 0

Part (a)

Vide (4), P(Z < 2) – P(Z < 0) = (1 – 0.028) – (1 – 0.5) = 0.472 ANSWER 1

Part (b)

Vide (4) and (1), P(Z < 0) – P(Z < 0) = 0 ANSWER 2

Part (c)

Vide (4) and (3), P(Z < - 1.64) – P(Z < - 1.96) = (0.0505) – (0. 025) = 0.0255 ANSWER 3

Part (d)

Vide (4) and (3), P(Z < 0.82) – P(Z < - 0.82) = (1 - 0.2061) – (0.2061) = 0.4878 ANSWER 4

Part (e)

Vide (4), P(Z < 1.9) – P(Z < 0.5) = (1 – 0.0287) – (1 – 0.3085) = 0.2798 ANSWER 1

DONE