Question

Someone please help!

Data is collected on the masses, m kg, of 140 Humboldt penguins. (data in table) a) given that an estimate for the mean mass of the penguins is 4.09kg, find the value of k​

Answer 1

k = 3.94 (2 d.p.)

Explanation:

The given table is a grouped frequency table with continuous data (no gaps or overlaps between classes).

Mean of grouped data

`\displaystyle \text{Mean}=(\sum fx)/(\sum f)`

(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.

`\textsf{class mid-point }(x)= \frac{\textsf{lower class boundary} + \textsf{upper class boundary}}{2}`

Calculate the mid-points (x) of each class and fx:

`\begin{array} 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 & (3.5+k)/(2) & 36.75+10.5k \\\cline{1-4} k \leq m < 4.0 & 33 & (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}`

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:

`\implies 4.09=(466.25+27k)/(140)`

`\implies 572.6=466.25+27k`

`\implies 27k=106.35`

`\implies k=3.94\:\:(2\: \sf d.p.)`

Therefore, the value of k is 3.94 (2 d.p.)

Answer 2

• k = 6.26

Explanation:

Find the midpoint of each frequency range, multiply that number by the frequency, sum up and divide by the total frequency.

This will give the mean number.

The mean is:

• [17*( 3 + 3.5)/2 + 21*(3.5 + k)/2 + 33*(k + 4)/2 + 54*(4 + 4.5)/2 + 15*(4.5 + 6)/2]/140 = 4.09
• 55.25 + 10.5k + 36.75 + 16.5k + 66 + 229.5 + 78.75 = 140*4.09
• 17k + 466.25 = 572.6
• 17k = 572.6 - 466.25
• 17k = 106.35
• k = 106.35/17
• k = 6.26

Note, I expected the value of k between 3.5 and 4. This is a bit of illogical outcome.