# Powerpoint 7.1

```AP Statistics Section 7.1
Probability Distributions
Consider tossing two 4-sided dice.
What is the sample space?
1-1
1-2
1-3
1-4
2-1
2-2
2-3
2-4
3-1
3-2
3-3
3-4
4-1
4-2
4-3
4-4
We are often interested in
numerical outcomes in statistics.
Let X equal the sum of the two 4sided dice. What are the only
possible values of X?
2, 3, 4, 5, 6, 7, 8
Since the value of X will vary from
roll to roll, we call X a ________
random
variable.
A random variable is a variable
whose value is a numerical
outcome of some random
phenomenon.
A discrete random variable is a
random variable with a countable
number of possible outcomes.
The probability distribution of a discrete
random variable X lists the possible values
of X along with their probabilities.
Value of x
x1
Probability
p1
x2
p2
x3
…..
p3
…..
xk
pk
The probabilities must satisfy two requirements:
1. Every probability is a number between ___
0 and ___.
1
2. The sum of the probabilities is _____
1 (i.e p  p      p  1 )
1
2
k
Example: Construct the probability distribution
for the sum of two 4-sided dice.
X
2
3
4
5
6
7
8
P(X)
.0625
.125
.1875
.25
.1875
.125
.0625
.625
P(sum is less than 6) = _________
.9375
P(sum is at least 3) = __________
Example: What is the probability distribution of the
discrete random variable X that counts the number of
heads in four tosses of a coin?
HHHH
THHH
HTHH
HHTH
HHHT
P( X  3) 
TTHH
THTH
THHT
HTTH
HTHT
HHTT
________
.9375
HTTT
THTT
TTHT
TTTH
TTTT
P(1  X  4) 
x
P(x)
0
.0625
1
.25
2
.375
3
.25
4
.0625
________
.875
We can draw a probability histogram for the discrete
random variable’s distribution.
0 .4
Probability
0 .3
0 .2
0 .1
0 .0
0
1
2
3
4
What are possible values for the following two
scenarios?
Number of seniors at Bay High School
280, 281, 282,....
Average height of seniors at Bay High School
Between 66&quot; and 74' '
Could be 68.1 or 68.14 or 68.142
While the outcomes on the left are
discrete the outcomes on the right
are not because they are not
countable numbers but an entire
interval of numbers.
A continuous random variable is a
random variable which takes on all
values in an interval on numbers.
Because a continuous random variable
can take on any value in some interval,
it is impossible to talk about random
variables using the same model as we
did the discrete random variables
above. For continuous random
variables the probability distribution is
described by a ____________.
density curve
*The probability of any event will be the
area under the curve and above the values
of X that make up the event.
Since the sum of all the probabilities must
be __
1 the total area under the curve must
be __.
1
A very “straight forward” density
curve to work with is the uniform
distribution, where outcomes are
across the interval.
Example: The first BART train of the day is to arrive at
the Golden Gate Station at 07:30. Arrival times are
uniformly distributed between 07:29:45 and 07:30:30.
Find...
1
45
P(the train is early)
P( x  7 : 30)
7 : 30 : 30
7 : 29 : 45
(15)(1 / 45)  1 / 3
45 sec.
P(the train is between 10 and 20 seconds late)
P(7 : 30 : 10  x  7 : 30 : 20)
(10)( 1
)  10  2
45
45
9
P(the train arrives at exactly 0730)
(0)( 1
)0
45
A continuous distribution assigns a
probability of __0__ to each individual
outcome.
Hopefully you recall working with
normal distributions back in
Chapter 2. Normal distributions are
probability distributions.
Example: The random variable X represents the profit
made on a randomly selected day by a certain store.
Assume X is Normally distributed with a mean of \$360
and standard deviation \$50. (Our shorthand notation
for this is ________)
N (360,50) Find….
P ( X  400)
z
  360
  50
400  360
 .80
50
1  .7881  .2119
P ((250  X  450)
.9502
Normalcdf (400,100000,360,50)
Example: The random variable X represents the profit made on a
randomly selected day by a certain store. Assume X is Normally
distributed with a mean of \$360 and standard deviation \$50.
The probability is approximately 0.6 that on a randomly selected
day the store will make less than x0 amount of profit. Find x0 .
.6
  360
  50
For .6, z  .25
x - 360
.25 
50
x  372.5
invNorm(.6,360,50)  372.67
```