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On May 11, 2013 at 9:30 PM, the probability that moderate seismic activity (one moderate earthquake) would occur in the next 48 hours in Japan was about 1.08%. As in [link] , you bet that a moderate earthquake will occur in Japan during this period. If you win the bet, you win $100. If you lose the bet, you pay $10. Let X = the amount of profit from a bet. Find the mean and standard deviation of X .
x | P ( x ) | x ⋅ ( Px ) | ( x - μ ) 2 P ( x ) | |
---|---|---|---|---|
win | 100 | 0.0108 | 1.08 | [100 – (–8.812)] 2 ⋅ 0.0108 = 127.8726 |
loss | –10 | 0.9892 | –9.892 | [–10 – (–8.812)] 2 ⋅ 0.9892 = 1.3961 |
Mean = Expected Value = μ = 1.08 + (–9.892) = –8.812
If you make this bet many times under the same conditions, your long term outcome will be an average loss of $8.81 per bet.
Some of the more common discrete probability functions are binomial, geometric, hypergeometric, and Poisson. Most elementary courses do not cover the geometric, hypergeometric, and Poisson. Your instructor will let you know if he or she wishes to cover these distributions.
A probability distribution function is a pattern. You try to fit a probability problem into a pattern or distribution in order to perform the necessary calculations. These distributions are tools to make solving probability problems easier. Each distribution has its own special characteristics. Learning the characteristics enables you to distinguish among the different distributions.
Class Catalogue at the Florida State University. Available online at https://apps.oti.fsu.edu/RegistrarCourseLookup/SearchFormLegacy (accessed May 15, 2013).
“World Earthquakes: Live Earthquake News and Highlights,” World Earthquakes, 2012. http://www.world-earthquakes.com/index.php?option=ethq_prediction (accessed May 15, 2013).
The expected value, or mean, of a discrete random variable predicts the long-term results of a statistical experiment that has been repeated many times. The standard deviation of a probability distribution is used to measure the variability of possible outcomes.
Mean or Expected Value:
Standard Deviation:
Complete the expected value table.
x | P ( x ) | x * P ( x ) |
---|---|---|
0 | 0.2 | |
1 | 0.2 | |
2 | 0.4 | |
3 | 0.2 |
Find the expected value from the expected value table.
x | P ( x ) | x * P ( x ) |
---|---|---|
2 | 0.1 | 2(0.1) = 0.2 |
4 | 0.3 | 4(0.3) = 1.2 |
6 | 0.4 | 6(0.4) = 2.4 |
8 | 0.2 | 8(0.2) = 1.6 |
0.2 + 1.2 + 2.4 + 1.6 = 5.4
Find the standard deviation.
x | P ( x ) | x * P ( x ) | ( x – μ ) 2 P ( x ) |
---|---|---|---|
2 | 0.1 | 2(0.1) = 0.2 | (2–5.4) 2 (0.1) = 1.156 |
4 | 0.3 | 4(0.3) = 1.2 | (4–5.4) 2 (0.3) = 0.588 |
6 | 0.4 | 6(0.4) = 2.4 | (6–5.4) 2 (0.4) = 0.144 |
8 | 0.2 | 8(0.2) = 1.6 | (8–5.4) 2 (0.2) = 1.352 |
Identify the mistake in the probability distribution table.
x | P ( x ) | x * P ( x ) |
---|---|---|
1 | 0.15 | 0.15 |
2 | 0.25 | 0.50 |
3 | 0.30 | 0.90 |
4 | 0.20 | 0.80 |
5 | 0.15 | 0.75 |
The values of P ( x ) do not sum to one.
Identify the mistake in the probability distribution table.
x | P ( x ) | x * P ( x ) |
---|---|---|
1 | 0.15 | 0.15 |
2 | 0.25 | 0.40 |
3 | 0.25 | 0.65 |
4 | 0.20 | 0.85 |
5 | 0.15 | 1 |
Use the following information to answer the next five exercises: A physics professor wants to know what percent of physics majors will spend the next several years doing post-graduate research. He has the following probability distribution.
x | P ( x ) | x * P ( x ) |
---|---|---|
1 | 0.35 | |
2 | 0.20 | |
3 | 0.15 | |
4 | ||
5 | 0.10 | |
6 | 0.05 |
Define the random variable X .
Let X = the number of years a physics major will spend doing post-graduate research.
Define P ( x ), or the probability of x .
Find the probability that a physics major will do post-graduate research for four years. P ( x = 4) = _______
1 – 0.35 – 0.20 – 0.15 – 0.10 – 0.05 = 0.15
FInd the probability that a physics major will do post-graduate research for at most three years. P ( x ≤ 3) = _______
On average, how many years would you expect a physics major to spend doing post-graduate research?
1(0.35) + 2(0.20) + 3(0.15) + 4(0.15) + 5(0.10) + 6(0.05) = 0.35 + 0.40 + 0.45 + 0.60 + 0.50 + 0.30 = 2.6 years
Use the following information to answer the next seven exercises: A ballet instructor is interested in knowing what percent of each year's class will continue on to the next, so that she can plan what classes to offer. Over the years, she has established the following probability distribution.
Complete [link] using the data provided.
x | P ( x ) | x * P ( x ) |
---|---|---|
1 | 0.10 | |
2 | 0.05 | |
3 | 0.10 | |
4 | ||
5 | 0.30 | |
6 | 0.20 | |
7 | 0.10 |
In words, define the random variable X .
X is the number of years a student studies ballet with the teacher.
P ( x = 4) = _______
On average, how many years would you expect a child to study ballet with this teacher?
What does the column " P ( x )" sum to and why?
The sum of the probabilities sum to one because it is a probability distribution.
What does the column " x * P ( x )" sum to and why?
You are playing a game by drawing a card from a standard deck and replacing it. If the card is a face card, you win $30. If it is not a face card, you pay $2. There are 12 face cards in a deck of 52 cards. What is the expected value of playing the game?
You are playing a game by drawing a card from a standard deck and replacing it. If the card is a face card, you win $30. If it is not a face card, you pay $2. There are 12 face cards in a deck of 52 cards. Should you play the game?
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