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M. Continuous Distributions.

1. Introduction.

These are optional! 6.28-6.32 [6.29 – 6.33] (6.24 - 6.28)

2. Properties of the Normal Distribution.

Text 6.1-6.2, 6.4!, 6.5a-c [6.1, 6.2, 6.5a-e.] (6.1, 6.2, 6.5a-e.) M1 a-g, M2, M3.

3. Percentiles and Intervals about the Mean.

Text 6.5d, 6.6!, 6.8 [6.4, 6.5f-h, 6.8.] (6.4, 6.5f-h, 6.8.) M4, M1 h-j. Graded assignment 4. M8.

4. Normal Approximation to the Binomial Distribution.

6.41 [6.57*], M6, M7.

5. Normal Approximation to the Poisson Distribution.

? [6.60*], M5.

Exercise 6.60(Not in 8th edition): The number of cars arriving per minute at a toll booth is Poisson distributed with a mean of 2.5. What is the probability that in any given minute: a. No cars arrive.

b. Not more than 2 cars arrive? c. What is the approximate probability that in a ten minute period not more than 20 cars arrive? d. What is the approximate probability that in a ten minute period between 20 and 30 cars arrive?

This document includes only optional exponential distribution problems.

Exponential Distribution Problems

From the outline , when the mean time to a success is .

Note that this is only for . There is no probability below zero. The text uses in place of c. is mean arrivals per unit. This is a continuous distribution, so there is no distinction between ‘<’ and ‘’. The Instructor’s Solutions Manual provides the answers below.

Exercise 6.28 [6.29 in 9th] (6.24 in 8th edition):

(a) P(arrival time 0.1)

(b) P(arrival time > 0.1) = 1 – P(arrival time 0.1) = 1 – 0.6321 = 0.3679

= P(arrival time < 0.2) – P(arrival time < 0.1) = 0.8647 – 0.6321 = 0.2326

(d) P(arrival time < 0.1) + P(arrival time > 0.2) = 0.6321 + 0.1353 = 0.7674

Exercise 6.29 [6.30 in 9th] (6.25 in 8th edition):

(a) P(arrival time 0.1)

(b) P(arrival time > 0.1) = 1 – P(arrival time 0.1) = 1 – 0.9502 = 0.0498

= P(arrival time < 0.2) – P(arrival time < 0.1) = 0.9975 – 0.9502 = 0.0473

(d) P(arrival time < 0.1) + P(arrival time > 0.2) = 0.0025 + 0.1353 = 0.1378

Exercise 6.30 [6.31 in 9th] (6.26 in 8th edition):

(a) P(arrival time 0.4)

(b) P(arrival time > 0.4) = 1 – P(arrival time 0.4) = 1 – 0.9997 = 0.0003

= 0.99995 – 0.9997 = 0.00025

(d) P(arrival time < 0.4) + P(arrival time > 0.5) = 0.9997 + 0.000045 = 0.999745

251solnM0 04/05/03

Exercise 6.31 [6.32 in 9th] (6.27 in 8th edition): . The average number of arrivals per minute is 50, so the mean time to a success is 0.2 minutes.

(a) P(arrival time 0.05)

(b) P(arrival time 0.0167) = 1 – 0.4339 = 0.5661

P(arrival time 0.0167) = 0.6329

(d) If = 30, P(arrival time 0.05) = 0.7769

P(arrival time 0.0167) = 0.3941

Exercise 6.32 [6.33 in 9th] (6.28 in 8th edition):

(a)

(b) P(arrival time 5) = 0.99996

P(arrival time 5) = 0.9933