Statistics Assignment

Topics: Probability density function, Random variable, Variance Pages: 8 (2641 words) Published: February 4, 2013
Week 4 Exercises
Chapter 5 – Section 1. Question 5
To perform a certain type of blood analysis, lab technicians must perform two procedures. The first procedure requires either one or two separate steps, and the second procedure requires either one, two, or three steps. a. List the experimental outcomes associated with performing the blood analysis. Answer: There are two procedures that a lab technician must perform. The first procedure requires either one or two separate steps, which could be named as x1 and x2. The second procedure requires either one, two or three steps, which could be named as y1, y2 and y3. The experimental outcomes associated with performing the blood analysis from 1st procedure (x1,x2) and 2nd procedure: (y1,y2,y3): * (x1,y1), x1,y2,x1,y3

* x2,y1,x2,y2,x2,y3
b. If the random variable of interest is the total number of steps required to do the complete analysis (both procedures), show what value the random variable will assume for each of the experimental outcomes. Answer: Values for the random variable to be assumed for each of the outcomes. * (x1,y1)=2 => the 1st procedure required 1 step and the 2nd procedure required 1 step * x1,y2=3 => the 1st procedure required 1 step and the 2nd procedure required 2 steps * x1,y3=4 => the 1st procedure required 1 step and the 2nd procedure required 3 steps * x2,y1=3 => the 2nd procedure required 1 step and the 1st procedure required 1 step * x2,y2=4 => the 2nd procedure required 1 step and the 1st procedure required 2 steps * x2,y3=5 => the 2nd procedure required 1 step and the 1st procedure required 3 steps Chapter 5 – Section 2. Question 11

A technician services mailing machines at companies in the Phoenix area. Depending on the type of malfunction, the service call can take one, two, three, or four hours. The different types of malfunctions occur at about the same frequency. a. Develop a probability distribution for the duration of a service call Answer: A probability distribution for the duration of a service call, where x is the duration of the service call, and the probability of different malfunctions occur at the same frequency x| f(x)|

1| 0.25|
2| 0.25|
3| 0.25|
4| 0.25|
| 1.00|
b. Draw a graph of the probability distribution.
Answer: A graph of the probability distribution

c. Show that your probability distribution satisfies the conditions required for a discrete probability function. Answer: Required conditions for a discrete probability functions are fx≥0=>f1=f2=f3=f4=0.25≥0

Σfx=1=>f1+f2+f3+f4=0.25+0.25+0.25+0.25=1
d. What is the probability that a service call will take three hours? Answer: The probability that a service call will take three hours is 0.25 f3=0.25
e. A service call has just come in, but the type of malfunction is unknown. It is 3:00 pm and service technicians usually get off at 5:00 pm. What is the probability that the service technician will have to work overtime to fix the machine today? Answer: The probability that the service technician will have to work overtime to fix the machine today is 0.5. Since it is 3pm right now and technician stays overtime, then he might need three or four hours to fix the machine, because if it takes only one or two hour, the technician is done on time. Thus, the probability that it would take three or four hours: f3+f4=0.25+0.25=0.5

Chapter 5 – Section 2. Question 13
A psychologist determined that the number of sessions required to obtain the trust of a new patient is either 1, 2, or 3. Let x be a random variable indicating the number of sessions required to gain the patient’s trust. The following probability function has been proposed. fx=x6 for x=1,2 or 3

a. Is this probability function valid? Explain.
Answer: This probability function is valid, because it meets the required conditions for a discrete probability x≥0=>f1=16≥0
f2=26≥0
f3=36≥0
Σfx=1=>f1+f2+f3=16+26+36=1
b. What is the probability...
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