Averages of Statistical Data practice questions

• $\text{Mid-value} = \dfrac{\text{lower bound} + \text{upper bound}}{2}$
• $\text{Mid-values} = \dfrac{2+4}{2},\ \dfrac{4+6}{2},\ \dfrac{6+8}{2},\ \dfrac{8+10}{2}$
• $\text{Mid-values} = 3,\ 5,\ 7,\ 9$
• $\text{Estimated Mean} = \dfrac{\Sigma fx}{\Sigma f}$
• Multiply each mid-value by its frequency to get $fx$, then add the products.
• $\Sigma fx = 3 \times 14 + 5 \times 7 + 7 \times 13 + 9 \times 12$
• $= 42 + 35 + 91 + 108$
• $= 276$

Answer: $6$

• There are 8 values, so the median is the mean of the 4th and 5th values.
• The known values in order are $52,\ 53,\ 56,\ 80,\ 81,\ 82,\ 86$.
• Of the known values, $3$ are below $77$ and $4$ are above it. With 8 values, 4 lie below the median and 4 above it, so $x$ is below $77$.
• So the two middle values are $80$ and the larger of $56$ and $x$. They cannot be $56$ and $80$, whose mean is $68$, not $77$, so they are $x$ and $80$.
• $\dfrac{x + 80}{2} = 77$
• $x + 80 = 154$
• $x = 74$

Answer: $74$

• For the mode to remain 0, its frequency (13) must be strictly greater than all other frequencies.
• The frequency of value 3 is x, so we need $x < 13$.
• Since x must be a whole number, the greatest possible value is $x = 12$.

Answer: $12$