Calculate E(X) from given distribution

The probability distribution is explicitly provided in a table or formula, and the question asks to calculate E(X) directly using the standard formula.

15 questions · Easy -1.0

5.02b Expectation and variance: discrete random variables
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OCR S1 2007 June Q1
5 marks Easy -1.8
1 The table shows the probability distribution for a random variable X.
x0123
\(\mathrm { P } ( \mathrm { X } = \mathrm { x } )\)0.10.20.30.4
Calculate \(\mathrm { E } ( \mathrm { X } )\) and \(\operatorname { Var } ( \mathrm { X } )\).
OCR MEI S1 Q2
7 marks Easy -1.2
2 Two fair six-sided dice are thrown. The random variable \(X\) denotes the difference between the scores on the two dice. The table shows the probability distribution of \(X\).
\(r\)012345
\(\mathrm { P } ( X = r )\)\(\frac { 1 } { 6 }\)\(\frac { 5 } { 18 }\)\(\frac { 2 } { 9 }\)\(\frac { 1 } { 6 }\)\(\frac { 1 } { 9 }\)\(\frac { 1 } { 18 }\)
  1. Draw a vertical line chart to illustrate the probability distribution.
  2. Use a probability argument to show that
    (A) \(\mathrm { P } ( X = 1 ) = \frac { 5 } { 18 }\),
    (B) \(\mathrm { P } ( X = 0 ) = \frac { 1 } { 6 }\).
  3. Find the mean value of \(X\).
Edexcel S1 2004 January Q3
10 marks Easy -1.3
3. A discrete random variable \(X\) has the probability function shown in the table below.
\(x\)0123
\(\mathrm { P } ( X = x )\)\(\frac { 1 } { 3 }\)\(\frac { 1 } { 2 }\)\(\frac { 1 } { 12 }\)\(\frac { 1 } { 12 }\)
Find
  1. \(\mathrm { P } ( 1 < X \leq 3 )\),
  2. \(\mathrm { F } ( 2.6 )\),
  3. \(\mathrm { E } ( X )\),
  4. \(\mathrm { E } ( 2 X - 3 )\),
  5. \(\operatorname { Var } ( X )\)
Edexcel S1 2020 June Q1
5 marks Moderate -0.8
  1. The discrete random variable \(X\) takes the values \(- 1,2,3,4\) and 7 only.
Given that $$\mathrm { P } ( X = x ) = \frac { 8 - x } { k } \text { for } x = - 1,2,3,4 \text { and } 7$$ find the value of \(\mathrm { E } ( X )\)
Edexcel S1 2013 June Q7
18 marks Moderate -0.3
7. The score \(S\) when a spinner is spun has the following probability distribution.
\(s\)01245
\(\mathrm { P } ( S = s )\)0.20.20.10.30.2
  1. Find \(\mathrm { E } ( S )\).
  2. Show that \(\mathrm { E } \left( S ^ { 2 } \right) = 10.4\)
  3. Hence find \(\operatorname { Var } ( S )\).
  4. Find
    1. \(\mathrm { E } ( 5 S - 3 )\),
    2. \(\operatorname { Var } ( 5 S - 3 )\).
  5. Find \(\mathrm { P } ( 5 S - 3 > S + 3 )\) The spinner is spun twice.
    The score from the first spin is \(S _ { 1 }\) and the score from the second spin is \(S _ { 2 }\) The random variables \(S _ { 1 }\) and \(S _ { 2 }\) are independent and the random variable \(X = S _ { 1 } \times S _ { 2 }\)
  6. Show that \(\mathrm { P } \left( \left\{ S _ { 1 } = 1 \right\} \cap X < 5 \right) = 0.16\)
  7. Find \(\mathrm { P } ( X < 5 )\).
Edexcel S1 2017 June Q6
18 marks Moderate -0.8
6. The score, \(X\), for a biased spinner is given by the probability distribution
\(x\)036
\(\mathrm { P } ( X = x )\)\(\frac { 1 } { 12 }\)\(\frac { 2 } { 3 }\)\(\frac { 1 } { 4 }\)
Find
  1. \(\mathrm { E } ( X )\)
  2. \(\operatorname { Var } ( X )\) A biased coin has one face labelled 2 and the other face labelled 5 The score, \(Y\), when the coin is spun has $$\mathrm { P } ( Y = 5 ) = p \quad \text { and } \quad \mathrm { E } ( Y ) = 3$$
  3. Form a linear equation in \(p\) and show that \(p = \frac { 1 } { 3 }\)
  4. Write down the probability distribution of \(Y\). Sam plays a game with the spinner and the coin.
    Each is spun once and Sam calculates his score, \(S\), as follows $$\begin{aligned} & \text { if } X = 0 \text { then } S = Y ^ { 2 } \\ & \text { if } X \neq 0 \text { then } S = X Y \end{aligned}$$
  5. Show that \(\mathrm { P } ( S = 30 ) = \frac { 1 } { 12 }\)
  6. Find the probability distribution of \(S\).
  7. Find \(\mathrm { E } ( S )\). Charlotte also plays the game with the spinner and the coin.
    Each is spun once and Charlotte ignores the score on the coin and just uses \(X ^ { 2 }\) as her score. Sam and Charlotte each play the game a large number of times.
  8. State, giving a reason, which of Sam and Charlotte should achieve the higher total score.
    END
Edexcel S1 Q1
7 marks Easy -1.3
  1. The discrete random variable \(Y\) has the following probability distribution.
\(y\)\({ } ^ { - } 2\)\({ } ^ { - } 1\)012
\(\mathrm { P } ( Y = y )\)0.10.150.20.30.25
Find
  1. \(\mathrm { F } ( 0.5 )\),
  2. \(\mathrm { P } \left( { } ^ { - } 1 < Y < 1.9 \right)\),
  3. \(\mathrm { E } ( Y )\),
  4. \(\mathrm { E } ( 3 Y - 1 )\).
AQA S2 2011 January Q4
18 marks Moderate -0.8
4
  1. A red biased tetrahedral die is rolled. The number, \(X\), on the face on which it lands has the probability distribution given by
    \(\boldsymbol { x }\)1234
    \(\mathbf { P } ( \boldsymbol { X } = \boldsymbol { x } )\)0.20.10.40.3
    1. Calculate \(\mathrm { E } ( X )\) and \(\operatorname { Var } ( X )\).
    2. The red die is now rolled three times. The random variable \(S\) is the sum of the three numbers obtained. Find \(\mathrm { E } ( S )\) and \(\operatorname { Var } ( S )\).
  2. A blue biased tetrahedral die is rolled. The number, \(Y\), on the face on which it lands has the probability distribution given by $$\mathrm { P } ( Y = y ) = \begin{cases} \frac { y } { 20 } & y = 1,2 \text { and } 3 \\ \frac { 7 } { 10 } & y = 4 \end{cases}$$ The random variable \(T\) is the value obtained when the number on the face on which it lands is multiplied by 3 . Calculate \(\mathrm { E } ( T )\) and \(\operatorname { Var } ( T )\).
  3. Calculate:
    1. \(\mathrm { P } ( X > 1 )\);
    2. \(\mathrm { P } ( X + T \leqslant 9\) and \(X > 1 )\);
    3. \(\mathrm { P } ( X + T \leqslant 9 \mid X > 1 )\).
AQA S2 2006 June Q3
8 marks Easy -1.2
3 Morecrest football team always scores at least one goal but never scores more than four goals in each game. The number of goals, \(R\), scored in each game by the team can be modelled by the following probability distribution.
\(\boldsymbol { r }\)1234
\(\mathbf { P } ( \boldsymbol { R } = \boldsymbol { r } )\)\(\frac { 7 } { 16 }\)\(\frac { 5 } { 16 }\)\(\frac { 3 } { 16 }\)\(\frac { 1 } { 16 }\)
  1. Calculate exact values for the mean and variance of \(R\).
  2. Next season the team will play 32 games. They expect to win \(90 \%\) of the games in which they score at least three goals, half of the games in which they score exactly two goals and \(20 \%\) of the games in which they score exactly one goal. Find, for next season:
    1. the number of games in which they expect to score at least three goals;
    2. the number of games that they expect to win.
AQA S2 2008 June Q7
9 marks Easy -1.2
7
  1. The number of text messages, \(N\), sent by Peter each month on his mobile phone never exceeds 40. When \(0 \leqslant N \leqslant 10\), he is charged for 5 messages.
    When \(10 < N \leqslant 20\), he is charged for 15 messages.
    When \(20 < N \leqslant 30\), he is charged for 25 messages.
    When \(30 < N \leqslant 40\), he is charged for 35 messages.
    The number of text messages, \(Y\), that Peter is charged for each month has the following probability distribution:
    \(\boldsymbol { y }\)5152535
    \(\mathbf { P } ( \boldsymbol { Y } = \boldsymbol { y } )\)0.10.20.30.4
    1. Calculate the mean and the standard deviation of \(Y\).
    2. The Goodtime phone company makes a total charge for text messages, \(C\) pence, each month given by $$C = 10 Y + 5$$ Calculate \(\mathrm { E } ( C )\).
  2. The number of text messages, \(X\), sent by Joanne each month on her mobile phone is such that $$\mathrm { E } ( X ) = 8.35 \quad \text { and } \quad \mathrm { E } \left( X ^ { 2 } \right) = 75.25$$ The Newtime phone company makes a total charge for text messages, \(T\) pence, each month given by $$T = 0.4 X + 250$$ Calculate \(\operatorname { Var } ( T )\).
OCR MEI Further Statistics A AS 2018 June Q2
10 marks Moderate -0.8
2 In a quiz, competitors have to match 5 landmarks to the 5 British counties which the landmarks are in. The random variable \(X\) represents the number of correct matches that a competitor gets, assuming that the competitor guesses randomly. The probability distribution of \(X\) is given in the following table.
\(r\)012345
\(\mathrm { P } ( X = r )\)\(\frac { 11 } { 30 }\)\(\frac { 3 } { 8 }\)\(\frac { 1 } { 6 }\)\(\frac { 1 } { 12 }\)0\(\frac { 1 } { 120 }\)
  1. Explain why \(\mathrm { P } ( X = 4 )\) must be 0 .
  2. Explain how the value \(\frac { 1 } { 120 }\) for \(\mathrm { P } ( X = 5 )\) is calculated.
  3. Draw a graph to illustrate the distribution.
  4. Find each of the following.
    • \(\mathrm { E } ( X )\)
    • \(\operatorname { Var } ( X )\)
    • Find \(\mathrm { P } ( X > \mathrm { E } ( X ) )\).
    • There are 12 competitors in the quiz. Assuming that they all guess randomly, find the probability that at least one of them gets all five matches correct.
OCR MEI Further Statistics A AS 2021 November Q1
4 marks Easy -1.3
1 The random variable \(X\) represents the clutch size (the number of eggs laid) by female birds of a particular species. The probability distribution of \(X\) is given in the table.
\(r\)234567
\(\mathrm { P } ( X = r )\)0.030.070.270.490.130.01
  1. Find each of the following.
    • \(\mathrm { E } ( X )\)
    • \(\operatorname { Var } ( X )\)
    On average 65\% of eggs laid result in a young bird successfully leaving the nest.
    1. Find the mean number of young birds that successfully leave the nest.
    2. Find the standard deviation of the number of young birds that successfully leave the nest.
OCR MEI Further Statistics Major 2024 June Q1
5 marks Easy -1.8
1 The number of insurance policy sales made per month by a salesperson is modelled by the random variable \(X\), with probability distribution shown in the table.
\(r\)0123456
\(\mathrm { P } ( \mathrm { X } = \mathrm { r } )\)0.050.10.250.30.150.10.05
  1. Find each of the following.
    The salesperson is paid a basic salary of \(\pounds 1000\) per month plus \(\pounds 500\) for each policy that is sold.
  2. Find the mean and standard deviation of the salesperson's monthly salary.
OCR MEI S1 Q1
8 marks Moderate -0.8
Four letters are taken out of their envelopes for signing. Unfortunately they are replaced randomly, one in each envelope. The probability distribution for the number of letters, \(X\), which are now in the correct envelope is given in the following table.
\(r\)01
P(X = r)\(\frac{3}{8}\)\(\frac{1}{3}\)\(\frac{1}{4}\)0\(\frac{1}{24}\)
  1. Explain why the case \(X = 3\) is impossible. [1]
  2. Explain why P(\(X = 4\)) = \(\frac{1}{24}\). [2]
  3. Calculate E(\(X\)) and Var(\(X\)). [5]
Edexcel S3 Q3
7 marks Standard +0.3
The discrete random variable \(X\) has the probability distribution given below.
\(x\)247\(k\)
\(P(X = x)\)0.050.150.30.5
  1. Find the mean of \(X\) in terms of \(k\). [2 marks]
  2. Find the bias in using \((2\overline{X} - 5)\) as an estimator of \(k\). [3 marks]
Fifty observations of \(X\) were made giving a sample mean of 8.34 correct to 3 significant figures.
  1. Calculate an unbiased estimate of \(k\). [2 marks]