Menu
Sign up Sign in

AI SL · 13.2

Probability Distributions

0 of 45 skills done

0
13.2.2

Expected value of a discrete random variable

13.2.2

Expected value of a discrete random variable

Target: I can find the expected value E(X) from a probability distribution table. (+4 more)

By the end of this section

Key points

Expected value

  • \(E(X)=\sum x\,P(X=x)\): multiply each value by its probability, then add.
  • E(X) is the mean of X over a very large number of trials. It is also written μ.
  • E(X) need not be a value X can take. A fair die has E(X) = 3.5, but you can never roll 3.5.
  • E(X) is not the most likely value. The most likely value is the mode.
  • Link to data: a frequency table has mean \(\frac{\sum xf}{\sum f}\). Writing \(\frac{f}{\sum f}\) as a probability gives \(\sum x\,P(X=x)\).

Distributions from formulae and with unknowns

  • A formula such as \(P(Y=y)=0.025(y^2-7y+20)\) is a table in disguise. Substitute each listed value of y, write the table, and check that it adds to 1.
  • With an unknown k, use \(\sum P(X=x)=1\) first. Find k before anything else.
  • A probability must be between 0 and 1. If k has two solutions, reject the one that gives a negative or a too large probability.
  • For a conditional probability, divide by the probability of the condition, not by 1.

Counts without replacement

  • Build the distribution of the count first, using a tree or the number of ways to choose. Then use \(E(X)=\sum x\,P(X=x)\).
  • Check the answer: for n items taken from a group, the expected count is n times the proportion of the group that has the property.
  • If a context asks for a total over many trials, multiply E(X) by the number of trials.

Explore: a probability distribution table

Type the values X can take and the probability of each. Leave one probability empty and it is found for you.

x
P(X = x)
Total of the probabilities
Expected value, E(X)

Type a table to see its bars and its expected value.

  1. The marker under the axis is E(X). Change one probability at a time. Which way does the marker move, and why?
  2. Can E(X) be a value that X never takes? Try the fair die.
  3. In the game, change the prize so that the game is fair.

Worked examples

Problems like the ones in class. Pick one, try it, then open the working.
E(X) from a table

X is the number of toppings chosen on a pizza at a takeaway. The table shows its probability distribution. Find E(X).

x01234P(X = x)0.100.200.350.250.10
Working
  1. Multiply and add

    \(E(X)=0\times0.10 + 1\times0.20 + 2\times0.35 + 3\times0.25 + 4\times0.10=2.05\)

    Each value times its probability, then add.

  2. Link to a frequency table

    \(\text{200 pizzas: }f=20,40,70,50,20,\ \text{mean}=\frac{410}{200}=2.05\)

    If 200 pizzas follow this distribution, the usual mean of the data is the same number.

  3. What it means

    E(X) = 2.05

    No pizza has 2.05 toppings. E(X) is the long-run average, not a value X takes.

Video walkthrough coming
A distribution from a formula

The random variable Y has probability distribution \(P(Y=y)=0.025(y^2-7y+20)\) for y = 1, 2, 3, 4. Find E(Y).

Working
  1. Substitute each y

    \(P(Y=1)=0.025(1-7+20)=0.35\) \(P(Y=2)=0.025(4-14+20)=0.25\) \(P(Y=3)=0.025(9-21+20)=0.2\) \(P(Y=4)=0.025(16-28+20)=0.2\)

    Put y = 1, 2, 3, 4 into the formula.

  2. y1234P(Y = y)0.350.250.20.2
    Table

    \(0.35+0.25+0.2+0.2=1\)

    Always check the total before using the table.

  3. E(Y)

    \(E(Y)=1\times0.35 + 2\times0.25 + 3\times0.2 + 4\times0.2=2.25\)

    Multiply each y by its probability and add.

Video walkthrough coming
Expected count without replacement

A tin holds 9 shortbread biscuits and 3 ginger biscuits. Two biscuits are taken at random without replacement. X is the number of ginger biscuits taken. Find the probability distribution of X and E(X).

Working
  1. Probabilities

    \(P(X=0)=\frac{9}{12}\times\frac{8}{11}=\frac{6}{11},\ P(X=1)=2\times\frac{3}{12}\times\frac{9}{11}=\frac{9}{22},\ P(X=2)=\frac{3}{12}\times\frac{2}{11}=\frac{1}{22}\)

    Two orders give one ginger biscuit, so double that route.

  2. x012P(X = x)6/119/221/22
    Table

    \(\frac{6}{11}+\frac{9}{22}+\frac{1}{22}=1\)

    The probabilities add to 1: the table is complete.

  3. E(X)

    \(E(X)=0\times\frac{6}{11}+1\times\frac{9}{22}+2\times\frac{1}{22}=\frac{1}{2}=0.5\)

    Check: 2 biscuits, and 3 out of 12 are ginger: 2 × 0.25 = 0.5.

Video walkthrough coming
A table with k and 2k

The table shows the probability distribution of Y. (a) Find k. (b) Find P(Y < 5). (c) Find E(Y).

y2459P(Y = y)0.250.15k2k
Working
  1. Sum to 1

    \(0.25+0.15+k+2k=1\Rightarrow 3k=0.6\Rightarrow k=0.2\)

    Collect the k terms: k + 2k = 3k.

  2. P(Y < 5)

    \(P(Y<5)=P(Y=2)+P(Y=4)=0.25+0.15=0.4\)

    Y < 5 means y = 2 or y = 4. The value 5 is not included.

  3. E(Y)

    \(E(Y)=2\times0.25+4\times0.15+5\times0.2+9\times0.4=5.7\)

    Use the value of 2k, not k, for y = 9.

Video walkthrough coming