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Q II Answer the following (Any 2):

1A. A biologist is studying the distribution of heights (in centimeters) of a particular species of plant in a field. She measures the heights of 50 randomly selected plants and records the following data:

[tex]
\begin{array}{l}
157, 163, 152, 148, 155, 160, 158, 151, 154, 162 \\
150, 159, 156, 161, 153, 157, 162, 150, 155, 158 \\
164, 152, 157, 160, 156, 155, 162, 149, 154, 159 \\
161, 157, 153, 150, 158, 156, 155, 162, 160, 155 \\
153, 161, 159, 164, 158, 151, 152, 157, 155, 163
\end{array}
[/tex]

For the above data:
i. Construct a frequency distribution table.
ii. Calculate the mean.

Answer :

We are given a list of 50 plant heights (in centimeters):

[tex]$$
\begin{array}{cccccccccc}
157, & 163, & 152, & 148, & 155, & 160, & 158, & 151, & 154, & 162, \\
150, & 159, & 156, & 161, & 153, & 157, & 162, & 150, & 155, & 158, \\
164, & 152, & 157, & 160, & 156, & 155, & 162, & 149, & 154, & 159, \\
161, & 157, & 153, & 150, & 158, & 156, & 155, & 162, & 160, & 155, \\
153, & 161, & 159, & 164, & 158, & 151, & 152, & 157, & 155, & 163
\end{array}
$$[/tex]

We will solve the problem in two parts.

─────────────────────────────
Step 1. Construct the Frequency Distribution Table

To create the frequency table, count the number of times each distinct height appears. The counts are as follows:

[tex]\[
\begin{array}{c|c}
\textbf{Height (cm)} & \textbf{Frequency} \\ \hline
148 & 1 \\[6mm]
149 & 1 \\[6mm]
150 & 3 \\[6mm]
151 & 2 \\[6mm]
152 & 3 \\[6mm]
153 & 3 \\[6mm]
154 & 2 \\[6mm]
155 & 6 \\[6mm]
156 & 3 \\[6mm]
157 & 5 \\[6mm]
158 & 4 \\[6mm]
159 & 3 \\[6mm]
160 & 3 \\[6mm]
161 & 3 \\[6mm]
162 & 4 \\[6mm]
163 & 2 \\[6mm]
164 & 2 \\
\end{array}
\][/tex]

Let’s verify the total frequency:
[tex]$$
1+1+3+2+3+3+2+6+3+5+4+3+3+3+4+2+2 = 50.
$$[/tex]

─────────────────────────────
Step 2. Calculate the Mean

The mean [tex]$\bar{x}$[/tex] of the data is given by

[tex]$$
\bar{x} = \frac{\text{Sum of all heights}}{\text{Number of plants}}.
$$[/tex]

Since we have grouped data, we can compute the sum by multiplying each height by its frequency and then summing up all contributions. That is,

[tex]$$
\text{Sum} = 148(1) + 149(1) + 150(3) + 151(2) + 152(3) + 153(3) + 154(2) + 155(6) + 156(3) + 157(5)\\
\quad\quad\quad\quad + 158(4) + 159(3) + 160(3) + 161(3) + 162(4) + 163(2) + 164(2).
$$[/tex]

Let’s compute each term:

[tex]\[
\begin{array}{rcl}
148 \times 1 &=& 148, \\
149 \times 1 &=& 149, \\
150 \times 3 &=& 450, \\
151 \times 2 &=& 302, \\
152 \times 3 &=& 456, \\
153 \times 3 &=& 459, \\
154 \times 2 &=& 308, \\
155 \times 6 &=& 930, \\
156 \times 3 &=& 468, \\
157 \times 5 &=& 785, \\
158 \times 4 &=& 632, \\
159 \times 3 &=& 477, \\
160 \times 3 &=& 480, \\
161 \times 3 &=& 483, \\
162 \times 4 &=& 648, \\
163 \times 2 &=& 326, \\
164 \times 2 &=& 328.
\end{array}
\][/tex]

Now add all these:

[tex]$$
\begin{aligned}
\text{Sum} &= 148 + 149 + 450 + 302 + 456 + 459 + 308 + 930 + 468 + 785 \\
&\quad + 632 + 477 + 480 + 483 + 648 + 326 + 328 \\
&= 7829.
\end{aligned}
$$[/tex]

There are 50 measurements, so the mean is:

[tex]$$
\bar{x} = \frac{7829}{50} = 156.58 \, \text{cm (approximately)}.
$$[/tex]

─────────────────────────────
Final Answer

1. The frequency distribution table for the plant heights is:

[tex]$$
\begin{array}{c|c}
\textbf{Height (cm)} & \textbf{Frequency} \\ \hline
148 & 1 \\
149 & 1 \\
150 & 3 \\
151 & 2 \\
152 & 3 \\
153 & 3 \\
154 & 2 \\
155 & 6 \\
156 & 3 \\
157 & 5 \\
158 & 4 \\
159 & 3 \\
160 & 3 \\
161 & 3 \\
162 & 4 \\
163 & 2 \\
164 & 2 \\
\end{array}
$$[/tex]

2. The mean height of the plants is approximately

[tex]$$
\bar{x} \approx 156.58 \, \text{cm}.
$$[/tex]

This completes the solution.

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