Problem 9 - Algebra Form and Function

Question

A naturalist collects samples of a species of lizard and measures their lengths. Give the (a) sample size (b) mean (c) range (d) \(\quad\) standard deviation. $$ \begin{array}{l|c|c|c|c|c} \hline \text { Lizard no. } & 1 & 2 & 3 & 4 & 5 \\ \hline \text { Length }(\mathrm{cm}) & 5.8 & 6.8 & 6.9 & 6.9 & 7.0 \\ \hline \text { Lizard no. } & 6 & 7 & 8 & 9 & 10 \\ \hline \text { Length }(\mathrm{cm}) & 7.1 & 7.1 & 7.1 & 7.2 & 8.1 \\ \hline \end{array} $$

Step-by-Step Solution

Answer

Question: In a study, a naturalist collected samples of a species of lizard and measured their lengths. The lengths (in cm) of the lizards are: 5.8, 6.8, 6.9, 6.9, 7.0, 7.1, 7.1, 7.1, 7.2, and 8.1. Find the sample size, mean, range, and standard deviation of the given data set. Answer: The sample size is 10, the mean length is 6.99 cm, the range is 2.3 cm, and the standard deviation is 0.601 cm.

Step 1: Find the sample size

The number of lizards measured is the sample size. In this case, the naturalist measured 10 lizards. So, the sample size is 10.

Step 2: Calculate the mean

To calculate the mean, add all the lengths of the lizards and divide by the sample size. Mean = \(\frac{5.8 + 6.8 + 6.9 + 6.9 + 7.0 + 7.1 + 7.1 + 7.1 + 7.2 + 8.1}{10} = \frac{69.9}{10} = 6.99\,\text{cm}\)

Step 3: Calculate the range

The range is the difference between the highest and the lowest value in the data set. In this case, the range is: Range = Highest value - Lowest value = 8.1 - 5.8 = 2.3 cm.

Step 4: Calculate the standard deviation

To calculate the standard deviation, first, find the variance. Variance = \(\frac{\sum(X_i - \text{Mean})^2}{\text{Sample Size}}\) Variance = \(\frac{(5.8-6.99)^2 + (6.8-6.99)^2 + \cdots +(8.1-6.99)^2}{10}\) Once you have the variance, find the standard deviation. Standard Deviation = \(\sqrt{\text{Variance}}\) Calculate the variance: Variance = \(\frac{1.39^2 + 0.19^2 + 0.09^2 + 0.09^2 + 0.01^2 + 0.11^2 + 0.11^2 + 0.11^2 + 0.21^2 + 1.11^2}{10} = 0.361\) Now, calculate the standard deviation: Standard Deviation = \(\sqrt{0.361} = 0.601\,\text{cm}\) So, the final answers are: a) Sample size: 10 b) Mean: 6.99 cm c) Range: 2.3 cm d) Standard deviation: 0.601 cm

Key Concepts

Sample Size

In statistics, the sample size refers to the number of observations or data points collected in a study. It is essential for providing accurate and reliable results when analyzing data. For the lizard lengths, the sample size is determined by counting the number of lizards measured. This sample, consisting of 10 lizards, provides insights into the population of lizards. A larger sample size generally leads to more reliable and generalizable results.

When working with small sample sizes, there is often a higher chance of variability or slight inaccuracies in the results. However, in this example, the sample size of 10 is a decent starting point for analyzing patterns or distributions of lizard lengths.

Mean

The mean, often referred to as the average, is a measure of central tendency. It provides a single value that represents the center of a data set. To calculate the mean, add up all the individual data points, then divide by the number of data points.

For the lizard lengths:

• Sum of lengths: 69.9 cm
• Sample size: 10
• Mean: \( \frac{69.9}{10} = 6.99 \) cm

The mean offers a straightforward way to understand the general size of the lizards in the sample. It highlights that most lizards are around 6.99 cm in length, but doesn't account for variations within the data set.

Standard Deviation

Standard deviation is a key statistic that measures the amount of variability or spread in a set of data. It tells us how much individual data points differ from the mean. A larger standard deviation indicates more spread out data points, whereas a smaller standard deviation indicates data points that are close to the mean.

To find the standard deviation:

• First, calculate the variance by finding the average of the squared differences from the mean.
• Then, take the square root of the variance to get the standard deviation.
• For our lizard example, Variance = 0.361, so \( \text{Standard Deviation} = \sqrt{0.361} = 0.601 \) cm.

This value shows that while most lizard lengths are close to the mean of 6.99 cm, there is some variability in their sizes. Understanding standard deviation helps in grasping the consistency and reliability of the data.

Range

The range is one of the simplest measures of variability and indicates the span between the smallest and largest values in a data set. To determine the range, subtract the smallest data point from the largest one.

For the lizard lengths:

• Largest value: 8.1 cm
• Smallest value: 5.8 cm
• Range: \( 8.1 - 5.8 = 2.3 \) cm

The range provides a quick glimpse into the spread of the data. It shows that the lizard lengths vary by 2.3 cm, from the shortest to the longest. While useful for a quick overview, the range does not convey information about the distribution of values between these two extremes.