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Chapter 6 Continuous Random Variables

 

6.1 Uniform Random Variables

 

Definition of Uniform Distribution
A random variable X is said to be uniformly distributed in x (lying between a and b) if its density function is



And the distribution is called a uniform distribution.

The distribution function is given by

Distribution Function



Example:

Consider a variable x which is uniformly distributed between the values 2 and 5, then the density function is given by


Solution:

F(X) = P(X x) = (x - 2) / (5 - 2) = 1/3 (x - 2)


Now that we have an example solved, can you solve a problem which is similar to the example problem?

Problem:

What is the distribution function of a random variable x which is uniformly distributed between the values 3 and 9?

Mean and variance

 

Problem: Suppose we are interested in bidding on a piece of land and we know there is one other bidder. The seller has announced that the highest bid in excess of 10k YTL will be accepted. Assume that competitor’s bid x is random variable that is uniformly distributed between 10k YTL and 15k YTL.

a)       Suppose you bid 12k what is the probability that your bid will be accepted

b)       What amount should you bid to maximize the probability that you can get the property?

c)       Suppose you know someone who is willing to pay you 16k YTL for the property. Would you consider to bid less than the amount in part b? Why or why not?

           

 

6.2 Normal Random Variables (Z-Table)

 

 

PROBLEM AND ACTIVITY SET FOR CHAPTER SÝX - NORMAL DISTRIBUTION

In this section, you will learn:

1.       What is a continuous distribution - properties and finding probabilities.

2.       What is a normal distribution - properties and construction.

3.       Identify if a random variable follows a normal distribution or not.

4.       What is a standardized normal distribution and the relationship between the standardize normal random variable and any given normal random variable.

5.       How to use the normal table.

6.       Apply normal random variables in real world situations.

Terms: Normal random variable, Standardized normal random variable, Z, Bell-shape distribution.

Problems

1.       Let X be the SAT scores. Based on the previous record, the distribution shape of the SAT scores is symmetric about the mean, and most of scores are around the mean and a few scores are away from the mean.

Notation X ~ N(m, s). m measures the average SAT score , s measures the spread of SAT scores.

The probability is the area underneath the curve within the interval of interest.

For example, The proportion of SAT scores between 450 and 700 is denoted as P(450 < X < 700).

NOTE: For any continuous r.v.,

P(a < X < b) = P(a <= X < b) = P(a < X <= b) = P(a <= X < = b)

*** The Equality sign does not contribute to the probability in continuous case. ***

This says: P(X = a ) = 0.

Now, let's work on the following problems.

2. Previous record indicates that the population average SAT score is m = 450 and population s.d. , s = 100. And it is reasonable to assume the distribution shape of the SAT scores follow a normal distribution. That is X ~ N(450, 100).

    1. What is the proportion of SAT scores to be higher than 650?
    2. What is the probability of having an SAT score between 350 and 550?
    3. What is the probability of having an SAT score between 450 and 600?
    4. What is the probability of having an SAT score higher than 750?
    5. Find median, Q1 SAT scores.
    6. CMU will admit the top 60% of students. What is the minimum SAT scores for the admission to CMU?
    7. A student scored 720. She claimed it is in the top 5%. Is her claim correct?

3. Z~N(0,1). Find P(0 < Z > 1.28), P(Z > 1.28) , P(Z < -1.28)

4. X~N(-20,10) Then, Is it true for each of the following statements:

a.       P(X > -20) = .5,

b.       (b) P(X > 0) = P(X < 0),

c.       (c) P(X > -40) = P( X < 0),

d.       (d) P( X > -20) < P(X < 0)

5.       According to experience, the computer price follows approximately a normal distribution with average price $1200 and a s.d. $300.

a.       What will the probability that a PC will cost $1550 or higher?

b.       What is the 25th percentile PC price?

c.       Your PC costs you $900. What is the proportion of PCs having prices lower than yours?

d.       A computer price analyst indicated that the average PC price will be about 40th percentile of the current prices by the end of the year. What will be the average PC price by the end of the year based on this analyst?

e.       If a PC price is outside two s.d. of the average price, it is considered rare. Is a PC of $800 considered as a rare price?

6. Give three r. v. that reasonably follow normal distribution, and three variables which do not follow normal distributrion.

7.  Z ~ N(0,1).

Find P(0 < Z < 1.96), P(Z < 1.96), P( Z < -1.96), P(Z > 1.96), P(-1.96 < Z < 0)

Find zo so that P(0 < Z < zo) = .24

Find zo so that P(Z > zo) = .975

Find Q3

8.  X ~ (50, 20)

Find the corresponding z-value for X = -20, 0, 20, 50, 80, 110

Find P(20 < X < 60), P(10 < X < 40), P( X > 85), P( X < 60)

Find xo so that P(X < xo) = .025

Find xo so that P(20 < X < xo) = .75

 

9.   According to experience, the time for using computer per day for students follows approximately a normal distribution with average 100 minutes and a s.d. 40 minutes.

a.       What will the probability that a student will be on computer for more than 3 hours per day?

b.       What is the 25th percentile of the amount of time on computer for students?

c.       You are on computer for 30 minutes per day. What is the proportion of students who spend less time on computer than you?

d.       If the amount of time spent on computer is either in the top 1% or low 1% of user are considered rare, what will be the amount of time on computer to be considered rare on the higher end and lower end, respectively?

Normal Distribution

Consider some population where the numeric measurement of interest has mean and standard deviation. If these measurements are normally distributed then the following properties apply:

Properties:

1.       These measurements are symmetrically distributed about the mean.

2.       A histogram of these measurements will be bell shaped.

3.       These measurements satisfy the empirical rule:

1.       68.27% of the measurements will be within one standard deviation of the mean.

2.       95.45% of the measurements will be within two standard deviation of the mean.

3.       99.73% of the measurements will be within three standard deviation of the mean.

Problem:

Consider the population of starting salaries for DePaul CTI'2000 graduates. Assume that these salaries are normally distributed with mean $60,000 and standard deviation $2,000.

1.       Determine the proportion of graduates that earned between $56,000 and $64,000.

2.       Determine the proportion of graduates that earned more than $64,000.

3.       Determine the proportion of graduates that earned between $56,000 and $66,000.

Solution:

1.       Observe that $56,000 is $4,000 less than the mean and $64,000 is $4,000 more than the mean. That is we are interested in the proportion of salaries within two standard deviations of the mean and so the desired proportion is 95.45%.

2.       From the symmetry property we know that 50% of salaries are more than $60,000. Now, observe that $64,000 is $4,000 more than the mean. From the symmetry property and the empirical rule this means that (95.45/2)% of salaries will be between $60,000 and $64,000. Hence the desired proportion is 50%-(95.45/2)% which is 2.275%.

3.       Observe that $56,000 is $4,000 less than the mean and $66,000 is $6,000 more than the mean. From the symmetry property and the empirical rule this means that (95.45/2)% of salaries will be between $56,000 and $60,000. (99.73/2)% of salaries will be between $60,000 and $66,000. Hence the desired proportion is (99.73/2)%+(95.45/2)% which is 97.59%.

Standard Normal Distribution

Consider some population where the numeric measurement of interest is normally distributed with mean =0 and standard deviation =1. Such a normal distribution is referred to as a Standard Normal Distribution. Tables are available for the Standard Normal Distribution which allow us to determine the proportion of measurements between any two values to a reasonable degree of accuracy.

 

Work these problems for practice. 

               Question #1:
 
               Find the area under the standard normal curve:
 
Note that this could be rephrased "Find the proportion of observations
that satisfy the following if the distribution is standard normal".
 
               a) to the right of 1.25
 
   From the standard normal table we know that 78.87% of observations are
   between -1.25 and 1.25. Since the normal curve is symmetric, and we are
   only interested in the proportion of observations greater than 1.25, the
   desired proportion is 50 - (78.87)/2 = 10.565%
 
b) to the left of -0.4
 
   The argument is as above. In this case 31.08% of observations are
   between -0.4 and 0.4. Since we are interested in the proportion to the
   left of -0.4, the desired proportion is 50 - (31.08)/2 = 34.46%
 
c) to the left of 0.8
 
               This could be approached in several ways. The usual way we would solve
               this problem is to recognize that the proportion to the left of 0.8 is
               he sum of the proportion less than mu and the proportion between mu
               and 0.8. From the standard normal table, the proportion between -0.8
               and 0.8 is 57.63% so by symmetry the proportion between mu and 0.8 is
               57.63/2 hence the desired proportion is 50 + (57.63)/2 = 78.815%.
 
               Alternatively: (Note: skip unless you are curious to see another approach.)  
               Another way of solving this is to think of the proportion to the left of
                0.8 as the sum of the proportion between -0.8 and 0.8 and the proportion
                to the left of -0.8. The proportion between -0.8 and 0.8 is 57.63% and
               the proportion to the left of -0.8 is 50 - (57.63)/2 = 21.185%. Hence,
               using this approach, the desired proportion is also 78.815%.
 
               d) between 0.4 and 1.3
 
               This can also be worked several ways but this time I will present one
                solution. We can first compute the proportion between -1.3 and 1.3
               and then subtract the proportion between -0.4 and 0.4. This will give
                us the proportion between 0.4 and 1.3 plus the proportion between
               -1.3 and -0.4. We are only interested in the proportion between
                0.4 and 1.3 but, because of symmetry, we can simply divide by 2. So, 
                since the proportion between -1.3 and 1.3 is 80.64% and that between
               -0.4 and 0.4 is 31.08%, our desired proportion is (80.64 - 31.08)/2 = 
               24.78%
 
                Exercise: Solve using the approach discussed in class to see that you
                obtain the same answer (i.e. notice that you may subtract the proportion
                between mu and 0.4 from the proportion between mu and 1.3).
 
               e) between -0.3 and 0.9
 
               Again several possible approaches. I will outline one and leave it 
               up to you to complete the computation and present another. First,
               you could compute the proportion to the left of 0.9 and subtract the
               proportion to the left of -0.3. Alternatively, you could compute
               the proportion between 0 and -0.3 and add it to the proportion between
               0 and 0.9. The proportion between 0 and -0.3 is simply the proportion
               between -0.3 and 0.3 divided by 2. That is 23.58/2 = 11.79%. The
               proportion between 0 and 0.9 is the proportion between -0.9 and
               0.9 divided by 2. That is 63.19/2 = 31.595%. Hence the desired 
               proportion is 43.385%. 
 
               f) outside -1.5 to 1.5
 
               This is simply 100 minus the proportion between -1.5 and 1.5. That is
               100 - 86.64 = 13.36%.
 
 
               Standard Normal Distribution Problems
 
 
1.       Compute: P(-3.80 < z <-2.40)
 

Solution:

Compute: P(-3.80 < z <-2.40):
Ans.: .0081

 
2.       A normal distribution has a mean of 150 and a standard deviation of 15. An observation is

randomly drawn from this population. Find the probability of getting an observation between

171 and 179
 

Solution:         

 

 

ans.: .0540

 

3.       A normal distribution has a mean of 100 and a standard deviation of 20. What score cuts off the lower 25% of the distribution

Solution: A normal distribution has a mean of 100 and a standard deviation of 20. What score cuts off the lower 25% of the distribution.

Since this is a normal distribution, the plan of attack is to find the z-score corresponding to the lower 25% of the standard normal distribution and convert it to the required x. The z-score is approximately -0.67.