A number x is chosen random from the numbers − 4 −3,−2,−1,0,1,2,3,4 what is the probability that x 1

Usage notes and limitations:

• The data type (class) must be a built-in MATLAB® numeric type. For other classes, the static rand method is not invoked. For example, rand(sz,'myclass') does not invoke myclass.rand(sz).

• Size arguments must have a fixed size.

• See Variable-Sizing Restrictions for Code Generation of Toolbox Functions (MATLAB Coder).

  • If extrinsic calls are enabled and rand is not called from inside a parfor loop, generated MEX files use the same random number state as MATLAB in serial code. Otherwise, the generated MEX code and standalone code maintain their own random number state that is initialized to the same state as MATLAB.

This function fully supports thread-based environments. For more information, see Run MATLAB Functions in Thread-Based Environment.

Usage notes and limitations:

• The stream syntax rand(s,___) is not supported on a GPU.

• You can specify typename as 'gpuArray'. If you specify typename as 'gpuArray', the default underlying type of the array is double.

  To create a GPU array with underlying type datatype, specify the underlying type as an additional argument before typename. For example, X = rand(3,datatype,'gpuArray') creates a 3-by-3 GPU array of random numbers with underlying type datatype.

  You can specify the underlying type datatype as one of these options:

• You can also specify the numeric variable p as a gpuArray.

  If you specify p as a gpuArray, the underlying type of the returned array is the same as p.

For more information, see Run MATLAB Functions on a GPU (Parallel Computing Toolbox).

Usage notes and limitations:

• The stream syntax rand(s,___) is not supported for codistributed or distributed arrays.

• You can specify typename as 'codistributed' or 'distributed'. If you specify typename as 'codistributed' or 'distributed', the default underlying type of the returned array is double.

  To create a distributed or codistributed array with underlying type datatype, specify the underlying type as an additional argument before typename. For example, X = rand(3,datatype,'distributed') creates a 3-by-3 distributed matrix of random numbers with underlying type datatype.

  You can specify the underlying type datatype as one of these options:

• You can also specify p as a codistributed or distributed array.

  If you specify p as a codistributed or distributed array, the underlying type of the returned array is the same as p.

• For additional codistributed syntaxes, see rand (codistributed) (Parallel Computing Toolbox).

For more information, see Run MATLAB Functions with Distributed Arrays (Parallel Computing Toolbox).

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Two numbers $x$ and $y$ are chosen at random within the unit interval $[0, 1]$. What is the probability that the sum of these is greater than one and that the sum of their squares is less than one?

The following must be complied with: $x+y>1$ and $x^2+y^2<1$. We plot, and there is an area between these two. The problem I have is in calculating that area.

Learning Outcomes

The expected value is often referred to as the “long-term” average or mean. This means that over the long term of doing an experiment over and over, you would expect this average.

You toss a coin and record the result. What is the probability that the result is heads? If you flip a coin two times, does probability tell you that these flips will result in one heads and one tail? You might toss a fair coin ten times and record nine heads. Probability does not describe the short-term results of an experiment. It gives information about what can be expected in the long term. To demonstrate this, Karl Pearson once tossed a fair coin 24,000 times! He recorded the results of each toss, obtaining heads 12,012 times.
In his experiment, Pearson illustrated the Law of Large Numbers.

The Law of Large Numbers states that, as the number of trials in a probability experiment increases, the difference between the theoretical probability of an event and the relative frequency approaches zero (the theoretical probability and the relative frequency get closer and closer together). When evaluating the long-term results of statistical experiments, we often want to know the “average” outcome. This “long-term average” is known as the mean or expected value of the experiment and is denoted by the Greek letter μ. In other words, after conducting many trials of an experiment, you would expect this average value.

Note

To find the expected value or long term average, μ, simply multiply each value of the random variable by its probability and add the products.

A men’s soccer team plays soccer zero, one, or two days a week. The probability that they play zero days is 0.2, the probability that they play one day is 0.5, and the probability that they play two days is 0.3. Find the long-term average or expected value, μ, of the number of days per week the men’s soccer team plays soccer.

To do the problem, first let the random variable X = the number of days the men’s soccer team plays soccer per week. X takes on the values 0, 1, 2. Construct a PDF table adding a column x ⋅ P(x). In this column, you will multiply each x value by its probability.

Expected Value Table. This table is called an expected value table. The table helps you calculate the expected value or long-term average.

Add the last column x ⋅ P(x) to find the long term average or expected value: (0)(0.2) + (1)(0.5) + (2)(0.3) = 0 + 0.5 + 0.6 = 1.1.

The expected value is 1.1. The men’s soccer team would, on the average, expect to play soccer 1.1 days per week. The number 1.1 is the long-term average or expected value if the men’s soccer team plays soccer week after week after week. We say μ = 1.1.

Find the expected value of the number of times a newborn baby’s crying wakes its mother after midnight. The expected value is the expected number of times per week a newborn baby’s crying wakes its mother after midnight. Calculate the standard deviation of the variable as well.

You expect a newborn to wake its mother after midnight 2.1 times per week, on the average.

Add the values in the third column of the table to find the expected value of X:

μ = Expected Value = [latex]\displaystyle\frac{{105}}{{50}}[/latex] = 2.1

Use μ to complete the table. The fourth column of this table will provide the values you need to calculate the standard deviation. For each value x, multiply the square of its deviation by its probability. (Each deviation has the format x – μ).

Add the values in the fourth column of the table:

0.1764 + 0.2662 + 0.0046 + 0.1458 + 0.2888 + 0.1682 = 1.05

The standard deviation of X is the square root of this sum: [latex]\displaystyle \sigma = \sqrt{{1.05}} \simeq {1.0247}[/latex]

A hospital researcher is interested in the number of times the average post-op patient will ring the nurse during a 12-hour shift. For a random sample of 50 patients, the following information was obtained. What is the expected value?

The expected value is 2.24,

[latex]\displaystyle{0})\frac{{4}}{{50}}+{({1})}\frac{{4}}{{50}}+{({2})}\frac{{16}}{{50}}+{({3})}\frac{{14}}{{50}}+{({4})}\frac{{6}}{{50}}+{({5})}\frac{{2}}{{50}}={0}+\frac{{8}}{{50}}+\frac{{32}}{{50}}+\frac{{42}}{{50}}+\frac{{24}}{{50}}+\frac{{10}}{{50}}=\frac{{116}}{{50}}={2.32}[/latex]

Suppose you play a game of chance in which five numbers are chosen from 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. A computer randomly selects five numbers from zero to nine with replacement. You pay $2 to play and could profit $100,000 if you match all five numbers in order (you get your $2 back plus $100,000). Over the long term, what is your expected profit of playing the game?

To do this problem, set up an expected value table for the amount of money you can profit.

Let X = the amount of money you profit. The values of x are not 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. Since you are interested in your profit (or loss), the values of x are 100,000 dollars and −2 dollars.

To win, you must get all five numbers correct, in order. The probability of choosing one correct number is
[latex]\displaystyle\frac{{1}}{{10}}[/latex] because there are ten numbers. You may choose a number more than once. The probability of choosing all five numbers correctly and in order is

[latex]\displaystyle{(\frac{{1}}{{10}})}{(\frac{{1}}{{10}})}{(\frac{{1}}{{10}})}{(\frac{{1}}{{10}})}{(\frac{{1}}{{10}})}={({1})}{({10}^{{-{5}}})}={0.00001}[/latex]

Therefore, the probability of winning is 0.00001 and the probability of losing is [latex]\displaystyle{1}-{0.00001}={0.99999}[/latex]

The expected value table is as follows:

Αdd the last column. –1.99998 + 1 = –0.99998

Since –0.99998 is about –1, you would, on average, expect to lose approximately $1 for each game you play. However, each time you play, you either lose $2 or profit $100,000. The $1 is the average or expected LOSS per game after playing this game over and over.

You are playing a game of chance in which four cards are drawn from a standard deck of 52 cards. You guess the suit of each card before it is drawn. The cards are replaced in the deck on each draw. You pay $1 to play. If you guess the right suit every time, you get your money back and $256. What is your expected profit of playing the game over the long term?

Let X = the amount of money you profit. The x-values are –$1 and $256.

The probability of guessing the right suit each time is

[latex]\displaystyle{(\frac{{1}}{{4}})}{(\frac{{1}}{{4}})}{(\frac{{1}}{{4}})}{(\frac{{1}}{{4}})}=\frac{{1}}{{256}}={0.0039}[/latex]

The probability of losing is
[latex]\displaystyle{1}-\frac{{1}}{{256}}=\frac{{255}}{{256}}={0.9961}[/latex]

(0.0039)256 + (0.9961)(–1) = 0.9984 + (–0.9961) = 0.0023 or 0.23 cents.

Suppose you play a game with a biased coin. You play each game by tossing the coin once.
P(heads) = [latex]\displaystyle\frac{{2}}{{3}}[/latex]. If you toss a head, you pay $6. If you toss a tail, you win $10. If you play this game many times, will you come out ahead?

1. Define a random variable X.
2. Complete the following expected value table.
   

   	x	____	____
   WIN	10	[latex]\displaystyle \frac{{1}}{{3}} [/latex]	____
   LOSE	____	____	[latex]\displaystyle \frac{{-12}}{{3}} [/latex]

3. What is the expected value, μ? Do you come out ahead?

Solution:

1. X = amount of profit

2. 	x	P(x)	xP(x)
   WIN	10	[latex]\displaystyle \frac{{1}}{{3}} [/latex]	[latex]\displaystyle \frac{{10}}{{3}} [/latex]
   LOSE	–6	[latex]\displaystyle \frac{{2}}{{3}} [/latex]	[latex]\displaystyle \frac{{-12}}{{3}} [/latex]

3. Add the last column of the table. The expected value μ = [latex]\displaystyle \frac{{-2}}{{3}} [/latex]. You lose, on average, about 67 cents each time you play the game so you do not come out ahead.

Suppose you play a game with a spinner. You play each game by spinning the spinner once.
P(red) = [latex]\displaystyle\frac{{2}}{{5}}[/latex]. If you land on red, you pay $10. If you land on blue, you don’t pay or win anything. If you land on green, you win $10. Complete the following expected value table.

[latex]\displaystyle-\frac{{20}}{{5}}[/latex]  [latex]\displaystyle\frac{{2}}{{5}}[/latex]

Here’s the completed table:

Like data, probability distributions have standard deviations. To calculate the standard deviation (σ) of a probability distribution, find each deviation from its expected value, square it, multiply it by its probability, add the products, and take the square root. To understand how to do the calculation, look at the table for the number of days per week a men’s soccer team plays soccer. To find the standard deviation, add the entries in the column labeled (x – μ)2P(x) and take the square root.

Add the last column in the table. 0.242 + 0.005 + 0.243 = 0.490. The standard deviation is the square root of 0.49, or σ = [latex]\displaystyle\sqrt{{0.49}}[/latex] = 0.7

Generally for probability distributions, we use a calculator or a computer to calculate μ and σ to reduce roundoff error. For some probability distributions, there are short-cut formulas for calculating μ and σ.

Toss a fair, six-sided die twice. Let X = the number of faces that show an even number. Construct a table like the one in the Try It above and calculate the mean μ and standard deviation σ of X.

Solution:

Tossing one fair six-sided die twice has the same sample space as tossing two fair six-sided dice. The sample space has 36 outcomes:

Use the sample space to complete the following table:

Calculating μ and σ.

Add the values in the third column to find the expected value: [latex]\displaystyle \mu =\frac{{36}}{{36}}={1}[/latex]. Use this value to complete the fourth column.

Add the values in the fourth column and take the square root of the sum: [latex]\displaystyle \sigma =\sqrt {{\frac{{18}}{{36}}}} \simeq {0.7071}[/latex].

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 Iran was about 21.42%. Suppose you make a bet that a moderate earthquake will occur in Iran during this period. If you win the bet, you win $50. If you lose the bet, you pay $20. Let X = the amount of profit from a bet.

P(win) = P(one moderate earthquake will occur) = 21.42%

P(loss) = P(one moderate earthquake will not occur) = 100% – 21.42%

If you bet many times, will you come out ahead? Explain your answer in a complete sentence using numbers. What is the standard deviation of
X? Construct a table similar to the one in Example 5 to help you answer these questions.

Solution:

Mean = Expected Value = 10.71 + (–15.716) = –5.006.

If you make this bet many times under the same conditions, your long term outcome will be an average
loss of $5.01 per bet.

[latex]\displaystyle \text{Standard Deviation} = \sqrt{{{648.0964}+{176.6636}}} \simeq {28.7186}[/latex]

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 Example 6, 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.

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.

[latex]\displaystyle \text{Standard Deviation} = \sqrt{{{127.7826}+{1.3961}}} \simeq {11.3696}[/latex]

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.

References

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).

Concept Review

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.

Formula Review

Mean or Expected Value:
[latex]\displaystyle\mu=\sum_{{{x}in{X}}}{x}{P}{({x})}[/latex]

Standard Deviation:
[latex]\displaystyle\sigma=\sqrt{{\sum_{{{x}in{X}}}{({x}-\mu)}^{{2}}{P}{({x})}}}[/latex]