Cumulative Probabilities Example

Assume we have a random variable X. Cumulative probabilities provide, for each value x, the probability of a result less than or equal to X, P[ X <= x ]. ( <= is "less than or equal to." Don't blame me. . .call Netscape & Microsoft.)

Example 1

Here's the probability distribution for a discrete random variable X

x

f(x)
1 0.1
2 0.2
3 0.4
4 0.3

The cumulative distribution function tables, for each value x = 1, 2, 3, 4, the probability of a result less than or equal to x. (In efficient English "less than or equal to" is often written "at most" or "no more than." These three phrases have the same meaning.) For example

  • P[ X <= 1 ] = 0.1
  • P[ X <= 2 ] = 0.1 + 0.2 = 0.3
  • P[ X <= 3 ] = 0.1 + 0.2 + 0.4 = 0.7
  • P[ X <= 4 ] = 0.1 + 0.2 + 0.4 + 0.3 = 1

These probabilities can be tabled

x P[ X <= x ]
1 0.1
2 0.3
3 0.7
4 1.0

Example 2

Suppose that P[ X <= 3 ] = 0.65, P[ X <= 4 ] = 0.80 and no values between 3 and 4 are possible. Then P[ X = 4 ] = 0.80 - 0.65 = 0.15. So, to compute f(x), take

f(x) = P[ X <= x ] - P[ X <= x* ]

where x* is the largest possible value below x. If x is already the smallest value then

f(x) = P[ X <=x ].

Suppose the number of finish flaws on an automobile has the following cumulative probabilities.

x P[ X <= x ]
0 0.30119
1 0.66263
2 0.87949
3 0.96623
4 0.99225
5 0.99850
6 0.99975
7 0.99996
8 1.00000

Questions

Answers in red!

  1. Find the probability there are at most 2 flaws, P[ X <= 2 ].
    0.87949
  2. Find the probability there are exactly 2 flaws, P[ X = 2 ].
    0.87949 - 0.66263 = 0.21686
  3. Find the probability there are less than 2 flaws, P[ X < 2 ].
    0.66263
  4. Find the probability there are more than 2 flaws, P[ X > 2 ].
    1 - 0.87949 = 0.12051
  5. Find the probability there are at least 2 flaws, P[ X >= 2 ].
    1 - 0.66263 = 0.33737
  6. Find the probability there are no flaws.
    P[ X = 0 ] = P[ X <= 0] = 0.30119
  7. Find the probability there is at least one flaw.
    1 - 0.30119 = 0.69881
  8. Find the probability there are between 1 and 3 flaws, inclusive, P[ 1 <= X <= 3 ].
    0.96623 - 0.30119 = 0.66504

It is far better to understand what information is in the table-what the table "means"-and to think about the question, than it is to try to memorize 5 different strategies for handling these cases.There is an "easy" (if tedious) method for recovering f(x)-from each cumulative probability subtract its predecessor. For this example:

x P[ X <= x ]   f(x)
0 0.30119 0.30119 - 0.00000 = 0.30119
1 0.66263 0.66263 - 0.30119 = 0.36144
2 0.87949 0.87949 - 0.66263 = 0.21686
3 0.96623 0.96623 - 0.87949 = 0.08674
4 0.99225

0.99225 - 0.96623 =

 0.02602
5 0.99850 0.99850 - 0.99225 = 0.00625
6 0.99975 0.99975 - 0.99850 = 0.00125
7 0.99996 0.99996 - 0.99975 = 0.00021
8 1.00000 1.00000 - 0.99996 = 0.00004
      1.00000

You can use this table to find answers to the questions above. Whichever way you do it, your results should be the same.

Do you understand it? If so, perhaps you'd like to try the worksheet on this topic?