Download Calculus 1 Lecture Notes - Madison Area Technical College

Calc 1 Lecture Notes
Section 5.7
Page 1 of 5
Section 5.7: Probability
Big idea:.
Big skill:.
Main Formulas:
1. Properties of a probability density function (pdf) f(x) on the complete range of possible
outcomes a  x  b:
a. Probability is always positive: f(x)  0 for a  x  b
b
b. All possible choices must sum up to 100%:
 f ( x)dx  1
a
d
c. Probability of a range of measurements/choices: P(c  X  d) =
 f ( x)dx
c
b
2. The mean (most likely value) is:    xf ( x)dx .
a
c
3. The median (half above, half below) is found by solving for c: 0.5   f ( x)dx .
a
Question: What does child-birthin’ have to do with probability distribution functions?
Answer: The number of kids you can expect of either gender is a precursor to the normal
distribution function.
Example: If you have 2 kids, what are the probabilities of getting zero, one, or two girls?

You can list all the possibilities this way: GG, GB, BG, BB, where the first letter
represents the gender of the first kid, and the second letter represents the gender of the second
kid.

You also can list the possibilities using a “probability tree.” Tracing through every
branch of the tree yields the same four choices: GG, GB, BG, BB.
Calc 1 Lecture Notes
Section 5.7
Page 2 of 5

There is only one possibility (out of 4) that leads to no girls, there are two possibilities
(out of 4) that lead to one girl, and only one possibility (out of 4) that leads to two girls.
1
o Thus, the possibility of 0 girls is  0.25
4
2
o The possibility of 1 girl is  0.50
4
1
o The possibility of 2 girls is  0.25
4

These probabilities are frequently represented on a bar chart (or histogram) like this:
Probability of Number of Girls Out of Two Children
Probability
0.60
0.50
0.40
0.30
0.20
0.10
0.00
-1
-0.5
0
0.5
1
1.5
2
2.5
3
Number of Girls

Notice what happens if you add up the areas of each of the three rectangles:
(0.25)(1) + (0.50)(1) + (0.25)(1) = 1; this can be interpreted as: the chance of getting zero, one,
or two girls is 100% (obviously). Connection to calculus: this sum is a midpoint approximation
to the integral of some as yet unknown function.

Notice that if you want the probability of zero or one girls, the answer is:
(0.25)(1) + (0.50)(1) = 0.75 (or 75%); this is an approximation of the integral from 0 to 1…

A cool connection between probability and algebra is that the coefficients of the power of
a binomial match the number of chances of getting a certain outcome. Notice the similarity of
squaring the binomial to the 4 possible outcomes from having two children:
2
 G  B    G  B  G  B   GG  GB  BG BB  1 G 2  2  GB  1 B 2

Remember that Pascal’s Triangle can be used to find the coefficients in the expansion of
a binomial. However, Pascal originally did not invent his triangle to speed calculation of the
powers of a binomial; he was a gambling fiend who wanted an edge at the casino…
Calc 1 Lecture Notes
Section 5.7
Page 3 of 5
Example: If you have 2 kids, what are the probabilities of getting zero through four girls?
4

 G  B   1 G 4  4  G 3 B  6  G 2 B 2  4  GB3  1 B 4


There are 1 + 4 + 6 + 4 + 1 = 16 total outcomes; there are always 2n total choices…
Breakdown per possibility:
1
 0.0625
o Possibility of 0 girls is
16
4
 0.25
o Possibility of 1 girl is
16
6
 0.375
o Possibility of 2 girls is
16
4
 0.25
o Possibility of 3 girls is
16
1
 0.0625
o Possibility of 4 girls is
16
Probability of Number of Girls Out of Four Children
0.40
Probability
0.30
0.20
0.10
0.00
-1
0
1
2
3
4
5
Number of Girls
Example: If you have 20 kids, what are the probabilities?
Probability of Number of Girls Out of 20 Children
Probability
0.2
0.15
0.1
0.05
0
-1
1
3
5
7
9
11
13
15
17
19
21
Number of Girls
Discrete Probability
Normal Distribution, Sigma = 2.264
As the number of children (or coin tosses or radioactive decays or any other random process)
tends to infinity, the histogram gets closer and closer to the Normal Distribution:

1
The Normal Distribution Function: f  x  
e
2
 x   2
2 2
Calc 1 Lecture Notes
Section 5.7
Page 4 of 5
In the Normal Distribution Function,
 = the mean, or average value, which corresponds to the maximum of the function
 = the standard deviation, which measures the spread of the distribution function

1
is the “normalization constant” which guarantees  f  x  dx  1
2 

Once you have a probability distribution function, you can make many statistical calculations:
The distribution function for IQ, where the average IQ is defined as 100 and the standard

1
deviation is 10, is f  x  
e
10 2
 x 1002
200
What is the probability that someone has an IQ between 90 and 110 (i.e., one standard deviation
above or below the mean)?
P  90  x  110  
110

f ( x)dx
90
110
 x 100 2

1

e 200 dx

10 2 90
1

 fnInt(e^(-(X-100) 2 / 200), X ,90,110)
10 2
 0.6827
 68.27%
Calc 1 Lecture Notes
Section 5.7
Page 5 of 5
What is the probability that someone has an IQ over 130 (i.e., 3 standard deviations above the
average)?
130

1
P  x  130   1   f ( x)dx 
2  70

130
 x 100 2


1
1
200

 1 
e
dx

2  10 2 70


 0.0013
 0.13%
Show that this distribution returns 100 as the mean:
b
   xf ( x)dx
a
1

10 2

1
10 2
1

10 2
100

10 2
 0  100
 100


xe

 x 100 2
200
Let u  x 
dx


  x  100  100  e
 x 100 2

200
dx


  x  100  e

 x 100 2
200

1
dx  100 
10 2

e

 x 100 2
200
dx

 e du  100 1
u
c
a

a
100
 x 100 2

 x 100 2
 100eu du
And u(-) = -
And u() = -


1
f ( x)dx 
e 200 dx

10 2 
1

fnInt(e^(-(X-100) 2 / 200), X , 1000,100)
10 2
 0.5
2
200
2

x

 100  1   x  100
du


dx
200
100
du
 x  100  100
dx
  x  100  e

Show that the median is also 100: i.e., show 0.5   f ( x)dx
c
 x  100 

200
dx  100
du u
e dx
dx