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.
[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.
