Questions
Claire rolls a six-sided die 16 times.
The scores obtained are shown in the following frequency table.
It is given that the mean score is 3.
(a) Find the value of p and the value of q .
Each of Claire’s scores is multiplied by 10 in order to determine the final score for a game she is playing.
(b) Write down the mean final score.
▶️Answer/Explanation
Detailed solution
(a) Find the Value of \( p \) and the Value of \( q \)
From the frequency table,
The total number of rolls is 16, so the sum of the frequencies must equal 16:
\[
p + q + 4 + 2 + 0 + 3 = 16
\]
\[
p + q + 9 = 16
\]
\[
p + q = 7 \quad (1)
\]
The mean score is given as 3. The mean is calculated as the total sum of (score × frequency) divided by the number of rolls:
\[
\text{Mean} = \frac{\sum (\text{score} \times \text{frequency})}{16} = 3
\]
So:
\[
\frac{(1 \cdot p) + (2 \cdot q) + (3 \cdot 4) + (4 \cdot 2) + (5 \cdot 0) + (6 \cdot 3)}{16} = 3
\]
Calculate the numerator:
\( 1 \cdot p = p \),
\( 2 \cdot q = 2q \),
\( 3 \cdot 4 = 12 \),
\( 4 \cdot 2 = 8 \),
\( 5 \cdot 0 = 0 \),
\( 6 \cdot 3 = 18 \),
Total: \( p + 2q + 12 + 8 + 0 + 18 = p + 2q + 38 \).
Set up the equation:
\[
\frac{p + 2q + 38}{16} = 3
\]
Multiply both sides by 16:
\[
p + 2q + 38 = 48
\]
\[
p + 2q = 10 \quad (2)
\]
Now solve the system of equations:
– (1) \( p + q = 7 \),
– (2) \( p + 2q = 10 \).
Subtract (1) from (2):
\[
(p + 2q) – (p + q) = 10 – 7
\]
\[
q = 3
\]
Substitute \( q = 3 \) into (1):
\[
p + 3 = 7
\]
\[
p = 4
\]
So, \( p = 4 \) and \( q = 3 \). Let’s verify:
– Total frequency: \( 4 + 3 + 4 + 2 + 0 + 3 = 16 \) (correct).
– Mean: \( \frac{(1 \cdot 4) + (2 \cdot 3) + (3 \cdot 4) + (4 \cdot 2) + (5 \cdot 0) + (6 \cdot 3)}{16} = \frac{4 + 6 + 12 + 8 + 0 + 18}{16} = \frac{48}{16} = 3 \) (correct).
(b) Write Down the Mean Final Score
Each score is multiplied by 10 to determine the final score. The mean final score is the mean of the original scores multiplied by 10:
\[
\text{Mean final score} = 3 \cdot 10 = 30
\]
This makes sense, as multiplying each score by a constant factor scales the mean by the same factor.
……………………Markscheme……………..
Ans: (a) attempt to form equation for the sum of frequencies=16 or mean=3
\(p+q+4+2+3=16(⟹p+q=7)\)
\(\frac{p+2q+12+8+18}{16}=3(⟹p+2q=10)\) OR \(\frac{p+2q+12+8+18}{9+p+q}=3(⟹2p+q=11)\)
attempt to eliminate one variable from their equations
\(p+2(7-p)+38=48\) OR \(2(7-q)+q=11\)
\(p=4\) and \(q=3\)
(b) mean final score = 30