Define "discrete data" to the person sitting next to you. Now define "random" and "variable"

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Presentation transcript:

Define "discrete data" to the person sitting next to you. Now define "random" and "variable"

Which of the following could be described as all three - discrete, random and variable? The score on a dice The number of days in a particular month The length of a song in the charts The number of songs on an album in the charts The number of days it is before it rains again

Now try Q1 and 2 page154

A Probability Distribution This is a table showing the probabilty of each outcome of an experiment. Consider Q2 from p154 - A fair die is thrown 4 times and the number of times it scores 6, Y, is noted. Write down all the possible values for y. Work out the probabilities for each one. Show this answer in table form. What should all your probabilities add up to? Can you think of a formula you can use to summarise these probabilities?

P(X = x) = 4 C x ( 5 / 6 ) 4-x ( 1 / 6 ) x x = 1, 2, 3, 4 X01234 P(X = x) Below is the example of a probability distribution Please complete with your values from earlier And here is an example of its probabilty function Do all the values add to 1? If not why not?

Exercise 8A Q 3 - 9

If a particular value of a discrete random variable, X, is x then P(X ≤ x) is written as F(x) This is known as a cumulative distribution function and can be written as a function or a table Here is the distribution from our previous example with the cumulative distribution for you to add below it. X01234 P(X = x) F(X)

X01234 P(X = x) F(X) Don't forget this would be less than 1 if we hadn't rounded to 2 dp Sometimes the function F(x) is defined in algebraic terms and so this row can be filled in directly from the function rather than adding the probabilities. You could even be given the function F(x) and work backwards to find the P(x). The following example shows this.

An example question involving a cumulative distribution function

Exercise 8B Q 1 - 8

The Expected Value of a Discrete Random Variable - - The Mean Consider the simple random variable - score on a six sided unbiased die Write down the probability distribution for this in a table and as a function Add a row for the cumulative distribution and write down its function F(x) What do you think will be the mean value of a score on a die. Throw a die 20 times and calculate the mean. As a class lets combine these scores to work out the mean of throws. Would you consider this mean to be more or less accurate than your first estimate? Why?

Was the mean value of the score on an unbiased die as you expected? Recall how you find the mean of a frequency distribution. We could have put our dice scores into a frequency table. If we had put the whole class scores in it would have looked something like this: ScoreFrequency If we throw the dice enough times the scores in this column should all be roughly the same, as the probability of scoring each value are identical. fx That would mean the score in this column look like this 1f 2f 3f 4f 5f 6f 21f 6f Where the value of f is determined by a sixth of the number of times the dice was thrown The mean is 21f / 6f = 3.5 As we saw in our experiment

1/61/6 Mean of a probability distribution ScoreP(X = x) px 1 The mean is 21 / 6 / 1 = 3.5 1/61/6 1/61/6 1/61/6 1/61/6 1/61/6 1/61/6 21 / 6 6/66/6 5/65/6 4/64/6 3/63/6 2/62/6 Realise the sum of this column is always 1 so the mean of a probability distribution can be calculated by simply adding this column The mean of a discrete random variable or probability distribution is usually referred to as the "expected value" and denoted by E(x)

Variance of a probability distribution 1/61/6 ScoreP(X = x) px 1 1/61/6 1/61/6 1/61/6 1/61/6 1/61/6 1/61/6 21 / 6 6/66/6 5/65/6 4/64/6 3/63/6 2/62/6 px 2 1/61/6 36 / 6 25 / 6 16 / 6 9/69/6 4/64/6 91 / 6 91 / 6 - ( 21 / 6 ) / 12 The variance of a discrete random variable or probability distribution can be caluculated by Var(x) = px 2 - [E(x)] 2 The text book refers to px 2 as E(x 2 )

An Example Question

An example where you can use the known mean of a distribution to find an unkown probability - working backwards

Exercises 8C Q and Exercise 8D Q 2 - 6

Consider a game now that uses two fair cubical dice and requires the total of their scores plus an additional 10. What would the mean score be? Could you write that using the notation we have learned in this chapter? How could you write the Variance of this score? What would it be? Now it gets even more complicated

12 / 36 ScoreP(Y = y) py 1 1 / 36 2 / 36 3 / 36 4 / 36 5 / 36 6 / / / / / / 36 py / / / / / / / / 36 4 / 36 3 / 36 2 / 36 1 / / / / / / / / / / / 36 Var(Y) = 1769 / 6 - (17) / 6

Let Y = X + W + 10 where w is that a different die is thrown. This is the same distribution as x but needs to be considered separately as it is a separate throw E(x+w+10) = 17 Var(x+w+10) = 5 5 / 6 X where x is the distribution shown earlier of throwing a fair cubical die E(x) = 3.5 Var(x) = 2 11 / 12 Compare the two distributions

Consider a game, Z, now that doubles a die score then adds another 10. What would the mean score be? Could you write that using the notation we have learned in this chapter? How could you write the Variance of this score? What would it be?

ScoreP(Z = z) pz 1 1/61/6 17 pz / / / 6 1/61/6 1/61/6 1/61/6 1/61/6 1/61/6 22 / 6 20 / 6 18 / 6 16 / 6 14 / / / / / / 6 Var(Z) = 902 / 3 - (17) / 3

Let Z = 2X + 10 E(2x+10) = 17 Var(2x+10) = 11 2 / 3 X where x is the distribution shown earlier of throwing a fair cubical die E(x) = 3.5 Var(x) = 2 11 / 12 Compare the two distributions

In Summary Now do read through the examples in your text book and do Exercise 8E

Discrete Uniform Distributions Our example with a fair cubical dice is also an example of a Uniform Distribution This is because the probabilities of scoring each number are equal - one sixth The formulae: E(X) = (n+1) / 2 and Var(X) = (n+1)(n-1) / 12 are perhaps more useful ways of calculating in these instances Now try Exercise 8F to practice these types of question