Given the data, calculate the standard deviation. Then, on graph paper, use a scale of 2cm to represent 1 on the x-axis and 2cm to represent 1 on the y-axis, draw a relative cumulative frequency curve. Use the graph to find the 80th percentile.

The formula for standard deviation is $\sigma = \sqrt{\frac{\sum x^2f}{\sum f} - \left(\frac{\sum xf}{\sum f}\right)^2}$.

The standard deviation is $\boxed{1.723}$.

From the row for $x=3$ in the second table, CF = 10 and RCF = 0.10.
Since RCF = CF / Total Frequency, we have:

• For $x=2$: RCF = 0.02. CF ($c$) = RCF $\times$ Total Frequency = $0.02 \times 100 = 2$.
• For $x=4$: CF = 16. RCF = CF / Total Frequency = $16 / 100 = 0.16$. (Note: The question's table shows 0.17, but 0.16 is consistent with a total frequency of 100).
• For $x=5$: CF = 42. RCF = CF / Total Frequency = $42 / 100 = 0.42$.
• For $x=7$: CF = 60. RCF = CF / Total Frequency = $60 / 100 = 0.60$.
• For $x=8$: CF = 60. RCF = CF / Total Frequency = $60 / 100 = 0.60$.

The completed table is:

Plot the points (upper class boundary, relative cumulative frequency). For discrete data, the upper class boundary for an $x$ value is $x+0.5$.
The points to plot are:

• $(1.5, 0)$ (starting point, assuming no fruits below $x=2$)
• $(2.5, 0.02)$
• $(3.5, 0.10)$
• $(4.5, 0.16)$
• $(5.5, 0.42)$
• $(7.5, 0.60)$
• $(8.5, 0.60)$

Use the given scales:

• X-axis: 2 cm to represent 1 unit for $0 \le x \le 8$.
• Y-axis: 2 cm to represent 0.1 for relative cumulative frequency. Draw a smooth curve connecting these points.

The 80th percentile corresponds to a relative cumulative frequency of 0.80.

On your drawn graph, find 0.80 on the y-axis. Draw a horizontal line from $y=0.80$ until it intersects the relative cumulative frequency curve. From the intersection point, draw a vertical line down to the x-axis. The value where this vertical line intersects the x-axis is the 80th percentile.
Since I cannot draw the graph, I cannot provide the numerical answer for the 80th percentile.