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
Question 6
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))
Calculating each term:
For x = 0: 0×121=0
For x = 3: 3×32=2
For x = 6: 6×41=1.5
Now, summing these values:
E(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))
Calculating each term:
For x = 0: 02×121=0
For x = 3: 32×32=6
For x = 6: $6^2 \times \frac{1}{4} = 9$$
Thus,
E(X2)=0+6+9=15
Now, we can calculate the variance:
Var(X)=E(X2)−(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×(1−p)+5×p=3
Expanding gives:
2−2p+5p=3
Combining terms:
2+3p=3
Thus,
3p=1⇒p=31
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:
y
2
5
P(Y = y)
2/3
1/3
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)
Using the probabilities:
P(X=6)=41,P(Y=5)=31
Thus,
P(S=30)=41×31=121
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:
S
0
2
6
12
15
30
P(S = s)
1/12
2/3 * (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))
You compute this based on the earlier results of the distribution:
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 X2 as her score, which is always greater or equal to Sam's score when X is non-zero. The expected value of E(X2) will generally exceed that of E(S) due to the nature of squaring larger values, thus leading to higher total scores. Therefore, Charlotte should achieve the higher score.