Transcript
Chapter 7: Continuous Distributions
Chapter Objectives
When you finish this chapter you should be able to
distinguish between discrete and continuous random variables.
state the parameters and uses of the uniform, normal and exponential distributions
select the appropriate continuous distribution in a problem context.
sketch uniform, normal and exponential probability density functions and areas.
use a table or spreadsheet to find uniform, normal or exponential areas for a given X.
solve for X for a given area from a uniform, normal or exponential model.
Quiz Yourself
True/False Questions
A continuous random variable X is uniformly distributed between 10 and 20 (inclusive), then the probability that X falls between 12 and 15 is 0.30.
All continuous probability distributions discussed in your text require the use of integral calculus to compute probability and the area under the curve.
Theoretically, the mean, median, and the mode are all equal for any normal distribution.
Any set of normally distributed data can be transformed to its standardized form.
Under a Standard Normal curve, the probability associated with the interval -1 to -2 is exactly the same as the probability associated with the interval 0 to 1.
A random variable X is standardized when each value of X has the mean of X subtracted from it, and the difference is divided by the variance of X.
If the mean of an exponential distribution is 2, then the value of the parameter ? is 2.0.
The mean and the variance of an exponential distribution are equal to each other.
Multiple Choice Questions
The probability density function, f(x), for any continuous random variable X, represents
A. all possible values that X will assume within some interval.
B. the probability that X takes on a specific value.
C. the area under the curve at a specific value.
D. the height of the function at a specific value.
Women’s heights are normally distributed with a mean of 65 inches and a standard deviation of 0.75 inches. Men’s heights are normally distributed with a mean of 69 inches and a standard deviation of 0.85 inches. Which is more extreme, a woman who is 5 feet tall or a man who is 6 feet tall?
A. The woman who is five feet tall.
B. The man who is six feet tall.
C. They are equally extreme.
D. There is no way to solve this problem.
Business analysts studying a portfolio which contains 30 stocks have decided that the daily returns on the portfolio are normally distributed with mean of 5.5% and a standard deviation of 2.5%. Use this information to answer the next THREE questions.
What proportion of the time will the returns on the portfolio be greater than 7%?
=NORMDIST(0.07,0.055,0.025,1)
=NORMDIST(7,38.5,15.4,1)
=1-NORMDIST(7,38.5,15.4,1)
=1-NORMDIST(0.07,0.055,0.025,1)
What proportion of the time will the portfolio have losses?
=1-NORMDIST(-0.01,0.055,0.0046,1)
=1-NORMDIST(0,0.055,0.025,1)
=NORMDIST(0,0.055,0.025,1)
=NORMDIST(-0.01, 0.055,0.0046,1)
What is the smallest return of the portfolios 10% most successful days?
=NORMINV(0.10,0.055,0.0046)
=NORMINV(0.9,0.055,0.025)
=1-NORMINV(0.9,0.055,0.0046)
=1-NORMINV(0.1,0.055,0.025)
President Dwight Eisenhower, in a major address to Congress, stated, “I was shocked to discover that fully 50% of the American population has an IQ that is below [the mean]! I challenge this Congress to address this problem and solve it!” As a statistician you know that IQ is normally distributed. What would you recommend to Congress?
A. Require all persons with an IQ less than the mean to return to public school for remedial instruction until the problem is resolved.
B. Prohibit anyone with an IQ greater than the mean from attending public schools until the problem is resolved.
C. Rescale the IQ assessment so that the mean is lower.
D. Rescale the IQ assessment so that the mean is higher.
E. None of the above.
The waiting time for a table at Nick’s on Friday night is a uniform random variable defined over the interval from 0 to 20 minutes. Use this information for the following THREE questions.
What is the probability that a randomly selected party will wait between 5 and 10 minutes to get a table?
A. 0.20 B. 0.25 C. 0.30 D. 0.05
What is the expected wait for a table at Nick’s on Friday nights?
A. 5 B. 7.5 C. 10 D. 12.5
What is the probability that a party will wait 5 minutes for a table?
A. 0.25 B. 0.05 C. 0.01 D. 0.00
Human gestation is a normally distributed random variable with a mean of 266 days and a standard deviation of 4.6 days. A birth that occurs in the lower 5% is considered premature and a birth that occurs in the upper 5% is considered postmature. What is the upper limit of gestation for premature infants and the lower limit for postmature infants?
A. 256 days and 275 days
B. 258 days and 274 days
C. 255 days and 272 days
D. 258 days and 271 days.
Like the normal distribution, the exponential density function f(x)
A. is bell-shaped.
B. is symmetrical.
C. approaches infinity as x approaches zero.
D. approaches zero as x approaches infinity.
Which of the following distributions is appropriate to measure the length of time between arrivals at a grocery checkout counter?
A. Uniform distribution
B. Normal distribution
C. Exponential distribution
D. Poisson distribution
Solved Problems from Text
7.2 a. C
b. D
c. C
7.4 For a continuous PDF, we use the area under the curve to measure the probability. The area above a single point is defined to be zero so if we summed up all the point probabilities we would have a sum equal to zero. In addition, by definition there are an infinite number of points in the interval over which a continuous random variable is defined.
7.6 a. P(X < 10) for U(0,50) = (10-0)/(50-0) = 0.2
b. P(X > 500) for U(0,1000) = (1000-500)/(1000-0) = 0.5
c. P(25 < X < 45) for U(15,65) = (45-25)/(65-15) = .4
7.8 a. ?= (2500+4500)/2 = 3500
b. ????= 577.3503
c. The first quartile is the midpoint between a and the median: 3500+2500)/2 = 3000. The third quartile is the midpoint between the edian and b: (4500+3500)/2 = 4000.
d. P(X < 3000) = P(2500 < X < 3000) = for U(2500,4500) =(3000- 500)/(4500-2500) =0.25.
e. P(X > 4000) = P(4000 < X <4500) = for U(2500,4500) = (4500- 00)/(4500-2500) = 0.25.
f. P(3000< X < 4000) = for U(2500,4500) =(4000-3000)/(4500-2500) =0.50.
7.10 a. The maximum height is 0?????. (Plug ? = 75 and ? =5 into the PDF.)
b. No, f(x) does not touch the X axis at any point. The distribution is asymptotic.
7.12 a. Yes
b. No, distribution could be skewed. Direction of skewness depends on how one defines years of education and which geographic region one is interested in.
c. No, distribution could be skewed right. Most bills will be delivered within a week but there may be a few that take much longer.
d. Yes, but there could be outliers.
7.16 a. NORMDIST(232000,232000,7000,TRUE) = 0.50
b. NORMDIST(239000,232000,7000,TRUE) ? NORMDIST(232000,232000,7000,TRUE) = 0.341345
c. NORMDIST(239000,232000,7000,TRUE) = 0.841345
d. NORMDIST(245000,232000,7000,TRUE) = 0.968355
e. 1 ? NORMDIST(225000,232000,7000,TRUE) = 0.84134474
7.18 Use Excel’s NORMINV function. NORMINV(0.975,3.3,.13) = 3.554795, NORMINV(.025,3.3,.13) = 3.045. The middle 95% is in the interval 3.045 to 3.555.
b. NORMDIST(3.50,3.3,.13,TRUE) = 0.061967919
7.20 Use Excel’s NORMINV function to give the X value associated with the cumulative probability.
a. NORMINV(.9,360,9) = 371.5339675
b. NORMINV(.5,360,9) = 360
c. NORMINV(.95,360,9) = 374.8036813
d. NORMINV(.2,360,9) = 352.4254106
e. NORMINV(.1,360,9) = 348.4660325
f. NORMINV(.25,360,9), NORMINV(.75,360,9) = 353.9295943, 366.0704057
g. NORMINV(.9,360,9) = 371.5339675
h. NORMINV(.025,360,9), NORMINV(.975,360,9) = 342.3603349, 377.6396651
i. NORMINV(.96,360,9) = 375.7561746
7.36 Use Excel’s =EXPONDIST(x, ?,1)
a. P(X > 30 minutes) = 1- EXPONDIST(0.50, 4.2,1) =0.122456
b. P(X < 15 minutes) = EXPONDIST(0.25,4.2,1) = 0.6500623
c. P(15 < X < 30) =EXPONDIST(.5,4.2,1)-EXPONDIST(.25,4.2,1) = 0.2274 813
Quiz Yourself Answers
True/False
Multiple Choice
1
T
6
F
1
D
6
E
11
D
2
F
7
F
2
A
7
B
12
C
3
T
8
T
3
D
8
C
4
T
4
C
9
D
5
F
5
B
10
B
