4) In the survey of Class G students, the weights (kg) of 50 students are as follows:

\begin{tabular}{cccccccc}
25 & 27 & 40 & 29 & 40 & 27 & 35 & 40 \\
29 & 38 & 35 & 25 & 29 & 35 & 40 & 35 \\
35 & 25 & 38 & 39 & 38 & 29 & 25 & 29 \\
27 & 29 & 40 & 27 & 29 & 35 & 27 & 29 \\
38 & 38 & 40 & 29 & 35 & 38 & 25 & 35 \\
35 & 40 & 29 & 27 & 35 & 25 & 40 & 27 \\
25 & 27 & & & & & &
\end{tabular}

Find:
i) Represent in a frequency distribution table.
ii) Mean [tex]$(x)$[/tex]
iii) Median (Md)
iv) Mode (Mo)
v) Upper Quartile (Q3)
vi) Lower Quartile (Q1)



Answer :

Certainly! Let's solve each part of the problem step-by-step:

### i) Siswete Series Table (Frequency Distribution)

We first tally the frequency of each distinct weight.

Weights: [tex]\[25, 27, 40, 29, 40, 27, 35, 40, 29, 38, 35, 25, 29, 35, 40, 35, 35, 25, 38, 39, 38, 29, 25, 29, 27, 29, 40, 27, 29, 35, 27, 29, 38, 38, 40, 29, 35, 38, 25, 35, 35, 40, 29, 27, 35, 25, 40, 27, 25, 27\][/tex]

Frequency of each weight:
[tex]\[ \begin{aligned} &\text{Weight (kg)} &\text{Frequency (f)} \\ &25 && 8 \\ &27 && 8 \\ &29 && 10 \\ &35 && 10 \\ &38 && 7 \\ &39 && 1 \\ &40 && 6 \\ \end{aligned} \][/tex]

So, the Siswete series table represented as a frequency distribution is:

[tex]\[ \begin{array}{|c|c|} \hline \text{Weight (kg)} & \text{Frequency (f)} \\ \hline 25 & 8 \\ \hline 27 & 8 \\ \hline 29 & 10 \\ \hline 35 & 10 \\ \hline 38 & 7 \\ \hline 39 & 1 \\ \hline 40 & 6 \\ \hline \end{array} \][/tex]

### ii) Mean \( (\bar{x}) \)

To find the mean, we use the formula:

[tex]\[ \bar{x} = \frac{\sum_{i=1}^n x_i}{n} \][/tex]

Sum of all weights:

[tex]\[ 25 \cdot 8 + 27 \cdot 8 + 29 \cdot 10 + 35 \cdot 10 + 38 \cdot 7 + 39 \cdot 1 + 40 \cdot 6 \][/tex]

[tex]\[ = 200 + 216 + 290 + 350 + 266 + 39 + 240 \][/tex]
[tex]\[ = 1601 \][/tex]

Number of weights \( n = 50 \).

Therefore, the mean weight:

[tex]\[ \bar{x} = \frac{1601}{50} = 32.02 \, \text{kg} \][/tex]

### iii) Median \( (Md) \)

To find the median, we need to order the data and find the middle value. Since we have 50 data points, the median will be the average of the 25th and 26th values in the ordered list.

Ordered weights:

[tex]\[ 25, 25, 25, 25, 25, 25, 25, 25, 27, 27, 27, 27, 27, 27, 27, 27, 29, 29, 29, 29, 29, 29, 29, 29, 29, 29, 35, 35, 35, 35, 35, 35, 35, 35, 35, 35, 38, 38, 38, 38, 38, 38, 38, 39, 40, 40, 40, 40, 40, 40 \][/tex]

The 25th and 26th values are both 29.

Thus, the median is:

[tex]\[ Md = \frac{29 + 29}{2} = 29 \, \text{kg} \][/tex]

### iv) Mode \( (mo) \)

The mode is the value that appears most frequently:

From our frequency distribution, weights 29 and 35 have the highest frequency of 10.

Thus, the mode is:

[tex]\[ mo = 29 \, \text{and} \, 35 \, \text{kg} \ (Bimodal) \][/tex]

### v) Upper Quartile (Q3)

The upper quartile (Q3) is the 75th percentile. For 50 data points, this corresponds to the value at position \( \frac{3(n + 1)}{4} = \frac{3(50 + 1)}{4} = 38.25 \). We take the average of the 38th and 39th values in the ordered list.

Ordered weights (from above), 38th value = 35, 39th value = 38.

Thus,

[tex]\[ Q3 = \frac{35 + 38}{2} = 36.5 \, \text{kg} \][/tex]

### vi) Lower Quartile (Q1)

The lower quartile (Q1) is the 25th percentile. For 50 data points, this corresponds to the value at position \( \frac{(n + 1)}{4} = \frac{(50 + 1)}{4} = 12.75 \). We take the average of the 12th and 13th values in the ordered list.

Ordered weights (from above), 12th value = 27, 13th value = 27.

Thus,

[tex]\[ Q1 = \frac{27 + 27}{2} = 27 \, \text{kg} \][/tex]

So, to summarize:

[tex]\[ \begin{aligned} \text{i) Siswete series Table:} & \\ \text{Weight (kg)} & \text{Frequency (f)} \\ 25 & 8 \\ 27 & 8 \\ 29 & 10 \\ 35 & 10 \\ 38 & 7 \\ 39 & 1 \\ 40 & 6 \\ \text{ii) Mean (x):} & \, 32.02 \, \text{kg} \\ \text{iii) Median (Md):} & \, 29 \, \text{kg} \\ \text{iv) Mode (mo):} & \, 29 \, \text{and} \, 35 \, \text{kg} \\ \text{v) Upper Quartile (Q3):} & \, 36.5 \, \text{kg} \\ \text{vi) Lower Quartile (Q1):} & \, 27 \, \text{kg} \\ \end{aligned} \][/tex]

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