Maths(四)---Probability and Statistics(9)

Standard Normal Distribution
例一.
Given Z is the normal distribution with μ = 0 and σ2 = 1, i.e. Z ~ N(0, 1), findthe following probabilities using the standard normal table.

(a) P(Z ≤ 1.25)
(b) P( Z >. 2.33)
(c) P(0.5 <. Z <. 1.5)
(d) P(Z <. -1.25)
(e) P(-1.5 <. Z <. -0.5)

Solution
(a) P(Z ≤ 1.25) = 0.8943

(b) P(Z > 2.33) = 1 – Φ(2.33) = 1 – 0.9901 = 0.0099

(c)P(0.5 <. Z <. 1.5) = Φ(1.5) – Φ(0.5)
= 0.9332 – 0.6915
= 0.2417

(d) P( Z <. -1.25) = P(z >. 1.25)= 1 – Φ(1.25)= 1 – 0.8944= 0.1056

(e)P(-1.5 <. Z <. -0.5) = P(0.5 <. Z <. 1.5) = 0.2417
例二.
Given Z is the normal distribution with μ = 0 and σ2 = 1, i.e. Z ~ N(0, 1), findthe value of c if
(a) P(Z ≤ c) = 0.8888
(b) P( Z > c) = 0.37
(c) P(Z <. c) = 0.025 (d) P(0 ≤ Z ≤ c) = 0.4924

Solution
(a) Φ(c) = 0.8888
c = 1.22

(b) 1 − Φ(c) = 0.37
Φ(c) = 0.63
c = 0.34

(c) Obviously c is negative and the standard normal table cannot be used directly.Recall that P(Z < -z) = P(Z > z) .
∴ 0.025 = P(Z <. c) = P(Z >. -c) = 1 – Φ(-c)
i.e. Φ(-c) = 0.975
-c = 1.96
c = -1.96

(d) P(0 ≤ Z ≤ c) = Φ(c) – 0.5 = 0.4924
Φ(c) = 0.992
c = 2.43

Standardization of Normal Random Variables
例一.


Suppose X ~ N(10, 6.25).
Find the following probabilities
(a) P(X <. 13)
(b) P(X > 5)
(c) P(8 <. X <. 15)

Solution

Let Z =(X -10)/2.5

(a) P(X <. 13) = P(Z <(13−10)/2.5)= P(Z <.1.2) = 0.8849

(b)P(X > 5) = P(Z >(5 −10)/2.5)= P(Z > -2) = P(Z <.2) = 0.9773

(c) P(8 <.X <.15)=P((8-10)/2.5<.Z<.(15-10)/2.5) = P(-0.8 <.Z <.2) = Φ(2) – Φ(-0.8) = Φ(2) – (1 – Φ(0.8))
= 0.9772 – (1 – 0.7881)
= 0.7653

例二.
The lifetime of light bulbs produced in a certain factory is normally distributed withmean 3000 hours and standard deviation 50 hours. Find the probability that arandomly chosen light bulb has a lifetime less than 2900 hours.

Solution
Let X be the lifetime of a randomly chosen light bulb. Then X ~ N(3000, 502).
Let Z =(X − 3000)/50
P(X <.2900) = P(Z <.-2) = 1 – 0.9772 = 0.0228


例三. If a random variable has the normal distribution with μ = 40 and σ = 2.4, find theprobabilities that it will take a value (a) less than 43.6 (b) greater than 38.2 (c) between 40.6 and 43.0 (d) between 35.8 and 44.2. Solution Let X be the random variable.


(a) P(X <.43.6) = P(Z <.(43.6− 40)/2.4)= P(Z <.1.5) = 0.9332

(b) P(X > 38.2) = P(Z >(38.2 − 40)/2.4) = P(Z > -0.75) = P(Z <.0.75) = 0.7734

(c) P(40.6 <.X <.43.0) = P((40.6 -40)/2.4<.Z<.(43.0-40)/2.4)

= P(0.25 <.Z <.1.25)

= 0.2957

(d)P(35.8 <.X <.44.2) = P((35.8-40)/2.4<.z<.(44.2-40)/2.4)

= P(-1.75 <.Z <.1.75)

= 0.9198