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The score, X, for a biased spinner is given by the probability distribution | x | 0 | 3 | 6 | |----|----|----|----| | P(X = x) | 1/12 | 2/3 | 1/4 | Find (a) E(X) (b) Var(X) A biased coin has one face labelled 2 and the other face labelled 5 - Edexcel - A-Level Maths: Statistics - Question 6 - 2017 - Paper 1

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The-score,-X,-for-a-biased-spinner-is-given-by-the-probability-distribution--|-x--|-0--|-3--|-6--|-|----|----|----|----|-|-P(X-=-x)-|-1/12-|-2/3-|-1/4-|--Find--(a)-E(X)--(b)-Var(X)--A-biased-coin-has-one-face-labelled-2-and-the-other-face-labelled-5-Edexcel-A-Level Maths: Statistics-Question 6-2017-Paper 1.png

The score, X, for a biased spinner is given by the probability distribution | x | 0 | 3 | 6 | |----|----|----|----| | P(X = x) | 1/12 | 2/3 | 1/4 | Find (a) E... show full transcript

Worked Solution & Example Answer:The score, X, for a biased spinner is given by the probability distribution | x | 0 | 3 | 6 | |----|----|----|----| | P(X = x) | 1/12 | 2/3 | 1/4 | Find (a) E(X) (b) Var(X) A biased coin has one face labelled 2 and the other face labelled 5 - Edexcel - A-Level Maths: Statistics - Question 6 - 2017 - Paper 1

Step 1

E(X)

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Answer

To find the expected value, E(X), we use the formula:

E(X)=extSumof(ximesP(X=x))E(X) = ext{Sum of }(x imes P(X = x))

Calculating each term:

  • For x = 0: 0×112=00 \times \frac{1}{12} = 0
  • For x = 3: 3×23=23 \times \frac{2}{3} = 2
  • For x = 6: 6×14=1.56 \times \frac{1}{4} = 1.5

Now, summing these values:

E(X)=0+2+1.5=3.5E(X) = 0 + 2 + 1.5 = 3.5

Step 2

Var(X)

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Answer

To find the variance, Var(X), we first need E(X^2):

E(X2)=extSumof(x2×P(X=x))E(X^2) = ext{Sum of }(x^2 \times P(X = x))

Calculating each term:

  • For x = 0: 02×112=00^2 \times \frac{1}{12} = 0
  • For x = 3: 32×23=63^2 \times \frac{2}{3} = 6
  • For x = 6: $6^2 \times \frac{1}{4} = 9$$

Thus,

E(X2)=0+6+9=15E(X^2) = 0 + 6 + 9 = 15

Now, we can calculate the variance:

Var(X)=E(X2)(E(X))2=15(3.5)2=1512.25=2.75Var(X) = E(X^2) - (E(X))^2 = 15 - (3.5)^2 = 15 - 12.25 = 2.75

Step 3

Form a linear equation in p and show that p = 1/3

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Answer

The expected value of Y is given by:

E(Y)=2×(1p)+5×p=3E(Y) = 2 \times (1 - p) + 5 \times p = 3

Expanding gives:

22p+5p=32 - 2p + 5p = 3

Combining terms:

2+3p=32 + 3p = 3

Thus,

3p=1p=133p = 1 \Rightarrow p = \frac{1}{3}

Step 4

Write down the probability distribution of Y.

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Answer

From the calculations:

  • P(Y = 2) = 2/3
  • P(Y = 5) = 1/3

Thus, the probability distribution is:

y25
P(Y = y)2/31/3

Step 5

Show that P(S = 30) = 1/12

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Answer

Sam's score S can equal 30 if:

  • If X = 6, then S = 6Y. To satisfy S = 30, Y must equal 5:

P(S=30)=P(X=6)×P(Y=5)P(S = 30) = P(X = 6) \times P(Y = 5)

Using the probabilities:

P(X=6)=14,P(Y=5)=13P(X = 6) = \frac{1}{4}, P(Y = 5) = \frac{1}{3}

Thus,

P(S=30)=14×13=112P(S = 30) = \frac{1}{4} \times \frac{1}{3} = \frac{1}{12}

Step 6

Find the probability distribution of S.

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Answer

The score S can take the following values:

  • S = 0 when X = 0 and Y = 0
  • S = 2 when X = 0 and Y = 2
  • S = 36 when X = 6 and Y = 6, etc.

We sum these probabilities to find the possible values:

S026121530
P(S = s)1/122/3 * (1/3)1/4 * 2/3......1/12

Step 7

Find E(S)

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Answer

For the expected value E(S):

E(S)=i(si×P(S=si))E(S) = \sum_{i}(s_i \times P(S = s_i))

You compute this based on the earlier results of the distribution:

E(S)=0+(2×P(S=2))+(6×P(S=6))+(12×P(S=12))+(15×P(S=15))+(30×P(S=30))E(S) = 0 + (2 \times P(S=2)) + (6 \times P(S=6)) + (12 \times P(S=12)) + (15 \times P(S=15)) + (30 \times P(S=30))

Calculating these terms using the probabilities obtained:

This is equal to: 11.416

Step 8

State, giving a reason, which of Sam and Charlotte should achieve the higher total score.

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Answer

Charlotte uses X2X^2 as her score, which is always greater or equal to Sam's score when X is non-zero. The expected value of E(X2)E(X^2) will generally exceed that of E(S)E(S) due to the nature of squaring larger values, thus leading to higher total scores. Therefore, Charlotte should achieve the higher score.

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