Transcript File - phs ap statistics

WARM – UP
#1. A company produces ceramic floor tiles which are
supposed to have a surface area of 16 square inches.
Due to variability in the manufacturing process, the actual
surface area of all tiles has a normal distribution with
mean 16.1 square inches and standard deviation 0.2
square inches. What is the proportion of tiles produced
by the process with surface area less than 16.0 square


inches?
16  16.1 

P(x < 16) = ?
P z


0.2

P(z < - 0.5) = Normalcdf(-E99, -0.5) = 0.3085 
1 
#2. The weight of a randomly selected can of a new soft
drink is known to have a normal distribution with mean
8.3 ounces and standard deviation 0.2 ounces. What
weight (in ounces) should be stamped on the can so that
only 2% of the cans are underweight?
?  8.3
2.05 
P(x < ?) = .02
0.2
z = InvNorm(.02) = -2.05
? = 7.89
1
RANDOM VARIABLES
A Random Variable is a variable whose value is a numerical
outcome of a random phenomenon. Usually denoted by X.
A Discrete Random Variable, X, has a finite or countable
number of possible values. The Probability Distribution of
X lists the values and their respective probabilities.
The following probability distribution represent the
AP Statistics scores from previous years.
AP Score
1
2
3
4
5
Probability 0.14 0.24 0.34 0.21 0.07
EXAMPLE:
1. All probabilities are: 0 ≤ p(x) ≤ 1
2. The sum of all probabilities equals one: Σ pi(x) = 1
A Continuous Random Variable, X, takes on all the infinite
numerical values within an interval. The Probability
Distribution of X is described by a Density curve. The
probability of any event is the area under the density curve.
UNIFORM Density Curve
Let X represent any random number between 0 and 5.
Find the following Probabilities:
EXAMPLE #1:
?
A= B·h
1 = 5·h
0
1
2
3
4
5
1. What is the Probability of any one number occurring?
0.24
P( 0.4 ≤ X ≤ 3.5) = 0.62
P(X > 2.2) = P(X ≥ 2.2) = 0.56
2. P( 0 ≤ X ≤ 1.2) =
3.
4.
1/5 = 0.2
NORMAL Density Curve
Let X represent any random number from the normal
distribution with mean 2.5 and standard deviation 0.8.
x
z
EXAMPLE #2:

1
.1
.9
0.0512
P( 0.4 ≤ X ≤ 3.5) = 0.8900
P(X > 2.2) = P(X ≥ 2.2) = 0.6462
1. P( 0 ≤ X ≤ 1.2) =
2.
3.
1.7 2.5 3.3 4.1 4.9
xU  2.5
xL  2.5
z
.8
.8
1
1
Normalcdf (zL, zU)
OR
Normalcdf (xL, xU, μ, σ)