Answer:
k = 3.94 (2 d.p.)
Step-by-step explanation:
The given table is a grouped frequency table with continuous data (no gaps or overlaps between classes).
Mean of grouped data
[tex]\displaystyle \text{Mean}=\dfrac{\sum fx}{\sum f}[/tex]
(where f is the frequency and x is the class mid-point).
To find an estimate of the mean, assume that every reading in a class takes the value of the class mid-point.
[tex]\textsf{class mid-point }(x)= \dfrac{\textsf{lower class boundary} + \textsf{upper class boundary}}{2}[/tex]
Calculate the mid-points (x) of each class and fx:
[tex]\begin{array}{| l | c | c | c |}\cline{1-4} \text{Mass, }m\:\text(kg) & \text{Frequency, }f & \text{Class mid-point, }x & fx \\\cline{1-4} 3 \leq m < 3.5 & 17 & 3.25 & 55.25\\\cline{1-4} 3.5 \leq m < k & 21 & \dfrac{3.5+k}{2} & 36.75+10.5k \\\cline{1-4} k \leq m < 4.0 & 33 & \dfrac{k+4.0}{2} & 16.5k+66\\\cline{1-4} 4.0 \leq m < 4.5 & 54 & 4.25 & 229.5 \\\cline{1-4} 4.5 \leq m < 6 & 15 & 5.25 & 78.75 \\\cline{1-4} \text{Totals} & 140 & & 466.25+27k\\\cline{1-4}\end{array}[/tex]
Given the mean is 4.09 kg, substitute the found values of f and fx (from the above table) into the mean formula and solve for k:
[tex]\implies 4.09=\dfrac{466.25+27k}{140}[/tex]
[tex]\implies 572.6=466.25+27k[/tex]
[tex]\implies 27k=106.35[/tex]
[tex]\implies k=3.94\:\:(2\: \sf d.p.)[/tex]
Therefore, the value of k is 3.94 (2 d.p.)
