Adam is measuring the heights in cm of his tomato plants.




Height (cm)
Frequency

$140
lt; h \leqslant 150$7

$150
lt; h \leqslant 160$10

$160
lt; h \leqslant 170$15

$170
lt; h \leqslant 180$19

$180
lt; h \leqslant 200$9





AFind the modal class.
BFind the mean and explain why this answer is an estimate.

Verified Answer :

  • The modal class is the interval with the highest frequency: \(\boxed{170 < h \le 180}\).
  • The estimated mean is calculated by summing the products of interval midpoints and frequencies, then dividing by the total number of plants: \(\frac{10075}{60} = \boxed{167.92\text{ cm}}\).
  • This mean is an estimate because we assume each plant's height within an interval is exactly the midpoint, which may not represent the actual distribution.

Explanation

  1. Analyze the frequency distribution

To find the mode and the mean of the heights of the tomato plants, we first analyze the given frequency distribution.
The distribution is as follows:

  • \(140 < h \le 150\): frequency \(7\)
  • \(150 < h \le 160\): frequency \(10\)
  • \(160 < h \le 170\): frequency \(15\)
  • \(170 < h \le 180\): frequency \(19\)
  • \(180 < h \le 200\): frequency \(9\)

The total number of plants measured is the sum of these frequencies: \(7 + 10 + 15 + 19 + 9 = 60\).

  1. Identify the modal class

The mode of a grouped frequency distribution is the modal class, which is the interval with the highest frequency. Looking at the data, the highest frequency is \(19\), which corresponds to the interval \(170 < h \le 180\). Therefore, the modal class is \(170 < h \le 180\).

  1. Calculate the estimated mean

To calculate the estimated mean for grouped data, we use the midpoint of each interval, denoted as \(x\), and multiply it by the corresponding frequency, \(f\).

  • For \(140 < h \le 150\): midpoint \(x = 145\), \(f \times x = 7 \times 145 = 1015\)
  • For \(150 < h \le 160\): midpoint \(x = 155\), \(f \times x = 10 \times 155 = 1550\)
  • For \(160 < h \le 170\): midpoint \(x = 165\), \(f \times x = 15 \times 165 = 2475\)
  • For \(170 < h \le 180\): midpoint \(x = 175\), \(f \times x = 19 \times 175 = 3325\)
  • For \(180 < h \le 200\): midpoint \(x = 190\), \(f \times x = 9 \times 190 = 1710\)

The sum of these products \(\sum(fx)\) is \(1015 + 1550 + 2475 + 3325 + 1710 = 10075\). The estimated mean is the total sum divided by the total number of plants: \(\text{Mean} = \frac{10075}{60} \approx 167.92\text{ cm}\). This is an estimate because the actual heights of the individual plants are not known, and we assume they are evenly distributed around the midpoint of each class interval.

Examples

This method is frequently used in biology and agriculture, for instance, when measuring crop yields or plant growth rates over a sample population. By grouping data into intervals, researchers can quickly summarize large datasets without needing the specific measurement of every single plant. This provides a practical snapshot for monitoring the health and development of the entire field.