Download 4.5 Continuous Random Variables

4.5 Continuous Random Variables
Modeling Distribution of a Continuous Random Variable. The probability distribution
of a continuous random variable, say x, is modeled by a function (curve) f(x) with the
property that
𝑏
P( a < x < b) = ∫𝑎 𝑓(𝑥 )𝑑𝑥 = area beneath the curve and between a and b (a < b)
1. The function f(x) is called a probability density function (pdf)
2. Total area under f(x) is
P(- ∞ < x < ∞) = 1
3. If x is a continuous random variable then P(x = a) = 0 for every a.
Consequently P( a < x < b) = P( a ≤ x < b) = P( a < x ≤ b) = P( a ≤ x ≤ b)
Exercise 1. A pdf of x is f(x) = x/2 if 0 < x < 2, and f(x) = 0 otherwise. Compute P( x ≤ 1)
4.6 Normal Distribution
A random variable has a normal distribution with the mean µ and the standard deviation
σ (σ >0) if its probability density function (pdf ) is
where π ≈ 3.141503, and e ≈ 2.718282.
Graph (bell-shaped curve)
µ-3σ
µ-2σ
µ-σ
µ
µ+σ
µ+2σ
µ+3σ
Exercise 2. Guess the mean and the standard
deviation for each of the following normal curves
Green
µ = ………….. σ = ………………..
Purple
µ = ………….. σ = ………………..
Orange
µ = ………….. σ = ………………..
The standard normal distribution is a normal distribution with mean µ = 0 and the
standard deviation σ = 1. A random variable with a standard normal distribution is called a
standard normal random variable, and is denoted z.
Computing normal probabilities.
A probability involving normal random variable is the proper area under a suitable normal curve and
can be found from Table II, calculator, or computer software. We will be using calculator (TI-83)
TI-83. Suppose that x has a normal distribution with mean µ and standard deviation σ
 normalcdf(a, b, µ, σ) = P(a < x < b) = area between a and b and under a normal curve with
mean µ and standard deviation σ. If a = - ∞, then use a = -10^99, if = - ∞, then use b = 10^99,
 invnorm(p, µ, σ) = the value x0 such that P(x < x0) = p [ pth percentile of x]
Both are in DISTR menu
Example. Let z be a standard normal random variable. Find the following
probabilities using TI-83.
1. [Ex. 4.16, p. 222] The probability that z falls between -1.33 and 1.33
P(-1.33 < z < 1.33) = normalcdf(-1.33, 1.33, 0,1) = .81648
2. [Ex. 4.17, p. 223] The probability that z exceeds 1.64
P( z > 1.64) = P (1.64 < z < ∞) = normalcdf(1.64, 10^99, 0,1) =.05050
3. [Ex. 4.18, p. 223] The probability that z lies to the left of .67
P( z < .67) = P (-∞ < z < .67) = normalcdf(-10^99,.67, 0,1) = .74857
4. [Ex. 4.19, p. 224] The probability that z exceeds 1.96 in absolute value.
P( |z| >1.96) = 2×P( z >1.96) = 2×P(1.96 < z < ∞) = .049996
Example. Assume that x has a normal distribution with the mean µ = 500, and the standard deviation
σ = 100. Find the probability that x falls between 450 and 600:
a. using standardization and Table II
b. using TI-83
SOLUTION
a. z-scores corresponding to a = 450 and b = 600 are
(450-500)/100 = -.50 and
(600-500)/100 = 1.00.
Hence P(450 < x <600) = P(-.50 < z < 1.00) =
= [area under standard normal curve between -.5 and 1] = 0.3413 + .1915 = .5328
b. P(450 < x <600) = normalcdf(450,600,500,100) =0.532807
Example. Find a value z0 of a standard normal variable z such that
P( z ≤ z0 ) = 0.90 (that is find the 90th percentile)
0.9
z0
z0 = invNorm(.9,0,1) = 1.28156
Example. Suppose that x is normally distributed with µ = 50 and σ = 3. Find a value x0 such that
P( x > x0 ) = 0.75 (that is find the 25th percentile)
x0 = invNorm(.25,50,3) = 47.98
0.25
0.75
Exercise 3. Let z be a standard normal random variable.
1. Find the following probabilities using TI-83. Draw the area that represents the probability.
a. P( z > -1.25)
b. P( z ≤ -.63)
c. P( -1.10 < z ≤ 1.63)
d. P( |z| > 1.59)
e. P( z = 2.00)
2. Find a value z0 such that. Illustrate the number of the graph of a normal curve.
a. P( z ≤ z0 ) = 0.15
b. P( z > z0 ) = 0.79
c. P( |z| ≤ z0 ) = 0.60
Exercise 4. Let x be normal random variable with µ = 30 and σ = 4.
1. Find the following probabilities using TI-83. Draw the area that represents the probability.
a. P (27 < x ≤ 41)
b. P( x > 25)
c. P( x ≤ 40)
d. P( x ≤ 1.59)
2. Find a value x0 such that. Illustrate the number of the graph of a normal curve.
a. P( x ≤ x0 ) = 0.25
b. P( x > x0 ) = 0.10
c. P(30 - x0 ≤ x ≤ 30+x0 ) = 0.60
Illustrate the numbers on the graphs of a proper normal curve.
Exercise 5 [4.59, p. 217]. Refer to Exercise 4.21 (p. 195) and the NHTSA crash test data for new cars.
One of the variables saved in the accompanying file is the severity of a driver's head injury when the
car is in a head-on collision with a fixed barrier while traveling at 35 miles per hour. The more points
assigned to the head-injury rating, the more severe the injury. The head-injury ratings can be shown to
be approximately normally distributed with a mean of 605 points and a standard deviation of 185
points. One of the crash-tested cars is randomly selected from the data, and the driver's head-injury
rating is observed.
a. Find the probability that the rating will fall between 500 and 700 points.
b. Find the probability that the rating will fall between 400 and 500 points.
c. Find the probability that the rating will be less than 850 points.
d. Find the probability that the rating will exceed 1,000 points.
Exercise 6 [like 4.109, p. 235]. Personnel tests are designed to test a job applicant's cognitive and/or
physical abilities. The Wonderlic IQ test is an example of the former; the Purdue Pegboard speed test
involving the arrangement of pegs on a peg board is an example of the latter. A particular dexterity test
is administered nationwide by a private testing service. It is known that for all tests administered last
year, the distribution of scores was approximately normal with mean 75 and standard deviation 7 .5.
The testing service reported to a particular employer that one of its job candidate's scores fell at the
95th percentile of the distribution . What was the candidate's score?