Math 365, Elementary Statistics 

Lesson 5 : Continuous Random Variables 5.1 Probability Density Function (pdf)

f(x)  =  1 if 0 ≤ x ≤ 1  
0 Otherwise 
Then we say X is uniformly distributed between 0 and 1 because it has the same density everywhere between 0 and 1.
Similarly, Y is said to be uniformly distributed between 1 and 3 if the pdf of Y is given by
g(x)  =  1/4 if 1 ≤ x ≤ 3  
0 Otherwise 
The Mean and Variance
The mean μ, variance σ^{2} and standard deviation σ of continuous random variables X are interpreted as we did for discrete random variables. As before, the mean μ, which is also called the expectation E(X), represents the average value of X.
But the definitions involve some calculus, which we are trying to avoid. If you have had calculus, I am giving the following definitions.
Suppose f(x) is the pdf of a continuous random variable X. Then the mean of X is
μ =E(X)=∫  ^{ ∞}  xf(x)dx 
_{}_{∞} 
and the variance of X is
σ ^{2} =Variance(X)=∫  ^{ ∞}  (x μ )^{2 }f(x)dx 
_{}_{∞} 
and the standard deviation σ is the square root of the variance σ^{2}.
Look at the following flash animations of graphs of some pdfs:
The most commonly encountered random variable in nature is the normal random variable. As we have seen in the last section, the probability distribution of a random variable is determined by the pdf of the random variable. The pdf of a normal random variable is described below.
PDF of a Normal Random Variable: Suppose f(x) is the pdf of a normal random variable X. Then we have the following properties of f(x).
Definition. A normal random variable is called a Standard Normal Random Variable if it has mean μ = 0 and standard deviation σ = 1. So, a N(0,1)random variable is called a standard normal variable. In some textbooks the standard normal random variable is denoted by Z. The GOOD NEWS is that a table is available to compute these probabilities. The following properties of Z will be useful.
Using the Probability Tables: Tables are used widely to compute probability. However, due to the use of various software programs on probability, the importance of such tables has declined. In this chapter, we will use the Ztable to compute probability for the standard normal random variable. We note the following:
Inverse Probability: Sometimes we will be given the probability and asked to compute a "cut off" point.
Given a N(μ,σ)random variable X, we can use the Ztable to compute probabilities for X because of the following theorem.
Theorem. Let X be a N(μ, σ)random variable. Then Z = [(Xμ)/(σ)] is a standard random varable. So,
P(a < X < b) = P(  a μ
σ 
< Z <  b μ
σ 
) 
P(a < X < b) = P(A < Z
< B)
where A= (aμ)/σ
and B= (bμ)/σ).
Now we can use the Ztable.
Problem Solving: We will have two types of problems in this section—probability computation and problems of inverse probability (or cutoff points).
Ubiquity of Normal Random Variables: Any random variable that we encounter in nature is, almost certainly, either normal or approximately normal. If there is one concept that you take from this course it is this: nature's random variables are normal or approximately normal. You will hear about normal random variables and the bell curve in your workplace or anywhere you may have to use statistics.
Problems on 5.2: the Normal Random Variable
Exercise 5.2.1. Let Z be the standard normal random variable.
Experiment with the normal
animation.
Solution
Exercise 5.2.2. Let X be a normal random variable with mean μ = 3 and standard deviation σ = 1.5 .
Experiment with the normal
animation.
Experiment with the Solution
Exercise 5.2.3. The length of life of some light bulbs produced in a factory is normally distributed with mean 8640 hours and standard deviation 1440 hours. Find the probability that a bulb will last
Exercise 5.2.4. The length X of a fish in
a lake has normal distribution with mean 67 cm and standard deviation
21 cm. What proportion (i.e, probability) of fish are between 44 cm
and 110 cm long?
Solution
Exercise 5.2.5. The diameter of the pumpkins
in my patch has normal distribution with mean 13 inches and standard
deviation 4.5 inches. What proportion (i.e., probability) of pumpkins
is above 22 inches?
Solution
Exercise 5.2.6. The annual expenditure X
of a student is approximately normally distributed with mean μ
= 11,000 dollars and standard deviation σ
= 1500 dollars. What percent of students spend less than 10,000 dollars?
Solution
Exercise 5.2.7. Suppose the annual production
X of milk per cow is normally distributed with μ
= 5500 liters and standard deviation σ
= 150 liters. What percent of cows have annual yield less than 5155
liters?
Solution
Exercise 5.2.8. The amount of vegetable oil
X produced by a machine in a day is normally distributed with μ
= 130 liters and standard deviation σ
= 25 liters. What is the probability that a machine will produce between
120 liters and 150 liters on a day?
Solution
Exercise 5.2.9. The weight X at birth of
babies is normally distributed with mean μ
= 114 oz and standard deviation σ =
18 oz. What percent of babies will have birth weight below 141 oz?
Solution
Problems on Cutoff values
Exercise 5.2.10. Let Z be the standard normal random variable.
Experiment with the normal
animation.
Solution
Exercise 5.2.11. The length X of a fish in
a lake has normal distribution with mean 67 cm and standard deviation
21 cm. On a fishing trip to the lake, you are instructed to release
those in the lower 33 percent in length. What is the cutoff length?
Solution
Exercise 5.2.12. The telephone company's
data shows that length X of their international calls has normal distribution
with mean 11.5 minutes and standard deviation 4.3 minutes. The company
decided to give a special rate for the longest 20 percent calls. What
is the cutoff time length?
Solution
Exercise 5.2.13. The weight X of babies (of
a fixed age) is normally distributed with with mean μ
= 212 oz and standard deviation σ =
25 oz. Doctors would be concerned (not necessarily alarmed) if a baby
is among the lower 5.05 percent in weight. Find the cutoff weight L
below which the doctors will be concerned.
Solution
Exercise 5.2.14. Monthly water consumption
X per household, in a subdivision in Kansas City, has normal distribution
with mean 15000 gallons and standard deviation 3000 gallons. It has
been decided that a surcharge will be imposed for those in the top 25
percent. Find the cutoff consumption U in gallons.
Solution
A wide range of random variables behave approximately like a normal random variable. One such example is binomial(n,p)random variables.
Roughly, if X is a B(n,p) random variable, then X behaves approximately like a normal random variable with mean μ = np and standard deviation σ = [np(1p)]^{1/2}.
As we know, a B(n,p) random variable X is discrete and
P(X=r) = _{n}C _{r}p^{r}(1p)^{nr} r=0,1,2,…,n.
On the other hand, if Y is a N(μ,
σ) random variable then
P(Y = r) = 0.
Because of this, some correction needs to be done. The following theorem
states how to use normal approximation to binomial random variables.
Theorem. Suppose X is a B(n,p) random variable.
If n is large and p is not very close to 0 or 1, then X behaves, approximately,
like a N(μ, σ)
random variable where
μ = np and standard deviation σ = [np(1p)]^{1/2}.
We have, for r=0,1,…,n
P(X = r) = P(r0.5 < X < r + .5) =P(L < Z
< R)
where L=(r0.5μ)/σ
and R=(r+0.5μ)/σ.
More generally, for r,s=0,1,…,n
P(r ≤ X ≤ s) = P(r0.5 < X < s +
.5) =P(L < Z < R)
where L=(r0.5μ)/σ
and R=(s+0.5μ)/σ.
Now use the Z table.
This adjustment by .5 on two sides is called continuity correction.
Problems on 5.3: Normal Approximation to Binomial
Exercise 5.3.1. A Lawrence bank knows that
35 percent of its customers will visit the drivethrough window. If
400 customers visit the bank, what is the approximate probability that
more than 120 will visit the drivethrough window?
Solution
Exercise 5.3.2. It is known that the probability that a household owns a food processor is 0.1. If 190 households are interviewed, find the approximate probability that
Exercise 5.3.3. The campaign committee of
a candidate claims that sixty percent of the voters are in favor of
the candidate. You interview 150 voters. Assuming that the campaign
committe's claim is accurate, what is the approximate probability that
less than 77 will favor the candidate?
Solution
Exercise 5.3.4. A technique is used to fertilize
eggs in a fertility clinic laboratory. It is known that the probability
that an egg will be fertilized by this technique is 0.1. If 500 eggs
are treated, what is the probability that at least 60 eggs will be fertilized?
Solution
Exercise 5.3.5. The probability that a computer
chip produced in a factory is defective is is .2. If you have a sample
of 60 chips, what is the probability that the number of defective chips
will be less than 20?
Solution
Exercise 5.3.6. The probability that a light
bulb produced by a machine is defective is p = 0.2. Suppose a quality
control inspector takes a sample of 120 bulbs. What is the probability
that more than 30 bulbs will be defective?
Solution
Exercise 5.3.7. Suppose the probability that
a student has access to the Internet is p = 0.8. Suppose you interview
160 students. What is the probability that less than 120 students will
have access to the Internet?
Solution
Exercise 5.3.8. Suppose that the probability
that a person favors medical use of marijuana is p = 0.6. If 780 individuals
are interviewed, what is the probability that less than 450 will be
in favor?
Solution
Exercise 5.3.9. Suppose that the probability
that a middleincome family invests in the stock market is p = 0.8.
If we interview 880 middleincome families, what is the probability
that more than 700 have invested in the stock market?
Solution
Exercise 5.3.10. Suppose that
an insurance company knows from experience that the probability that
a lifeinsurance policyholder will survive another 10 years is p = 0.9.
The company has 2280 policyholders. What is the probability that more
than 2025 will survive another 10 years.
Solution