Question 1
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 to determine the final score for a game she is playing.
(b) Write down the mean final score
▶️Answer/Explanation
(a) Since the die is rolled 16 times, the sum of all frequencies should be 16, that is:
\(p+q+4+2+0+3=16\)
\(p+q=7\)
\(p=7-q \dots (1)\)
Then, since the mean score is 3, so it follows:
\(\frac{(1\cdot p)+(2\cdot q)+(3\cdot 4) + (4\cdot 2)+(5\cdot 0)+(6\cdot 3)}{16}=3\)
\(p+2q+12+8+18=48\)
\(p+2q=10 \cdots (2)\)
Substitute (1) into (2):
\(7-q+2q=10\)
\(q=3\)
From (1):
\(p=7-3=4\)
(b) Since each of Claire’s scores is multiplied by 10, then the new frequency table should be:
| Score | Frequency |
| 10 | 4 |
| 20 | 3 |
| 30 | 4 |
| 40 | 2 |
| 50 | 0 |
| 60 | 3 |
Therefore, the mean final score is:
\(\frac{(10\cdot4)+(20\cdot3)+(30\cdot4)+(40\cdot2)+(50\cdot0)+(60\cdot3)}{16}=30\)
Question 2
It is given that \(\log_{10}{a}=\frac{1}{3}\), where \(a>0\).
Find the value of
(a) \(\log_{10}\left(\frac{1}{a}\right)\)
(b) \(\log_{1000}{a}\)
▶️Answer/Explanation
(a) Note that:
\(\log_{10}\left(\frac{1}{a}\right)=\log_{10}\left(a^{-1}\right)\)
\(\log_{10}\left(\frac{1}{a}\right)=-\log_{10}{a}\)
\(\log_{10}\left(\frac{1}{a}\right)=-\frac{1}{3}\)
(b) By using the change of base rule:
\(\log_{1000}{a}=\frac{\log_{10}{a}}{\log_{10}{1000}}\)
\(\log_{1000}{a}=\frac{\frac{1}{3}}{\log_{10}{10^{3}}}\)
\(\log_{1000}{a}=\frac{\frac{1}{3}}{3}\)
\(\log_{1000}{a}=\frac{1}{9}\)