Find Q $3$ Position: To find $Q3$ , we need to determine the position in the ordered data set. $Q3$ is the value below which $75\%$ of the data lies.
Calculate Total Livestock: First, calculate the total number of livestock by summing the frequencies: $4 + 8 + 10 + 12 + 6$ .
Calculate Position of Q $3$ : Total number of livestock = $4 + 8 + 10 + 12 + 6 = 40$ .
Find Value for Q $3$ Position: The position of Q $3$ is given by $(\frac{3}{4}) \times (\text{total number of livestock} + 1)$ . Let's calculate that.
Calculate Cumulative Frequencies: Position of $Q_3 = \left(\frac{3}{4}\right) \times (40 + 1) = \left(\frac{3}{4}\right) \times 41 = 30.75$ . So, we round up to the next whole number since we can't have a fraction of an observation.
Determine Interval for Q $3$ : Position of Q $3$ = $31$ . Now, we need to find the value that corresponds to this position in the cumulative frequency distribution.
Use Linear Interpolation: Let's calculate the cumulative frequencies: $4$ , $4+8=12$ , $12+10=22$ , $22+12=34$ , $34+6=40$ .
Calculate Q $3$ Value: The $31^{st}$ value falls in the interval $162-167$ , since the cumulative frequency just before this interval is $22$ and it is $34$ at the end of this interval.
Check for Mistakes: To find $Q3$ within the interval $162-167$ , we use linear interpolation. $Q3 = \text{lower limit} + \left[\frac{n/4 - F}{f}\right] \times \text{width of the interval}$ , where $F$ is the cumulative frequency before the interval, $f$ is the frequency of the interval, and $n$ is the total number of livestock.
Check for Mistakes: To find $Q_3$ within the interval $162-167$ , we use linear interpolation. $Q_3 = \text{lower limit} + \left[\frac{n/4 - F}{f}\right] * \text{width of the interval}$ , where $F$ is the cumulative frequency before the interval, $f$ is the frequency of the interval, and $n$ is the total number of livestock.$Q_3 = $162$ + \left[\frac{$31$ - $22$}{$12$}\right] * $5$ = $162$ + \left[\frac{$9$}{$12$}\right] * $5$ = $162$ + $0$.$75$ * $5$ = $162$ + $3$.$75$.
Check for Mistakes: To find $Q3$ within the interval $162-167$, we use linear interpolation. $Q3 = \text{lower limit} + \left[\frac{(n/4 - F)}{f}\right] * \text{width of the interval}$, where $F$ is the cumulative frequency before the interval, $f$ is the frequency of the interval, and $n$ is the total number of livestock.$Q3 = 162 + \left[\frac{(31 - 22)}{12}\right] * 5 = 162 + \left[\frac{9}{12}\right] * 5 = 162 + 0.75 * 5 = 162 + 3.75$. $Q3 = 162 + 3.75 = 165.75$. But this is not one of the options provided, so there must be a mistake.