Ans: (a) $3$ (b) No (c) (i) $42$ (ii) $11$ (iii) $10$
- The mean
$\begin{array}{cl}
= & \dfrac{1 \times15 + 2 \times 9 + \ldots + 7 \times 5}{15+9+\ldots + 5} \\
= & 3
\end{array}$ - Note that the mode is $1$ and the median is $2$. Therefore, the median and the mode are not equal.
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- Since the mean is increased by $1$, then
$\begin{array}{rcl}
\dfrac{1\times 15+2\times 9+\ldots + 5\times (4+n) + 6\times 2 + 7\times 5}{15+9+\ldots + 4 + n + 2 + 5} & = & 4 \\
\dfrac{126+5n}{42+n} & = & 4 \\
126 + 5n & = & 168 + 4n \\
n & = & 42
\end{array}$ - Since the median is increased by $2$, then the new median becomes $4$.
Since the number of children having the numbers of tokens less than $4$ is $26$, then there should be at least $27$ children having the numbers of token more than or equal to $4$.
$\begin{array}{rcl}
5+4+n+2+5 & = & 27 \\
n & = & 11
\end{array}$ - Since the mode remains unchanged, then the highest frequency is $15$ and the second highest frequency is $14$.
Then, the greatest value of $n$ is $10$.
- Since the mean is increased by $1$, then
