Problem 11
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. $$ \begin{array}{c|c|c|c|c|c|c|c} \hline \text { Length }(\mathrm{cm}) & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ \hline \text { No. lizards } & 1 & 6 & 26 & 36 & 23 & 6 & 2 \\ \hline \end{array} $$
Step-by-Step Solution
Verified Answer
Question: Based on the given data of the lengths of a species of lizard collected by a naturalist, find the following:
(a) Sample size
(b) Mean
(c) Range
(d) Standard deviation
Answer:
(a) Sample size: 100
(b) Mean: 7 cm
(c) Range: 6 cm
(d) Standard deviation: 1.11 cm (approximately)
1Step 1: Calculate the sample size
To find the sample size, we will sum up the number of lizards for each length, which can be found in the second row of the table.
Sample size = \(1 + 6 + 26 + 36 + 23 + 6 + 2 = 100\)
2Step 2: Find the mean
To find the mean, we will use the formula:
Mean = \(\frac{\sum (length * number \ of \ lizards)}{sample \ size}\)
Mean = \(\frac{4*1 + 5*6 + 6*26 + 7*36 + 8*23 + 9*6 + 10*2}{100}\)
Mean = \(\frac{4 + 30 + 156 + 252 + 184 + 54 + 20}{100}\)
Mean = \(\frac{700}{100}\)
Mean = \(7\)
3Step 3: Calculate the range
To find the range, we subtract the smallest length from the largest length:
Range = \(10 - 4 = 6\)
4Step 4: Determine the standard deviation
To find the standard deviation, we will use the following formula:
Standard deviation = \(\sqrt{\frac{\sum (length_{i} - mean)^{2} * number \ of \ lizards}{sample \ size}}\)
Standard deviation = \(\sqrt{\frac{(4-7)^{2}*1 + (5-7)^{2}*6 + (6-7)^{2}*26 + (7-7)^{2}*36 + (8-7)^{2}*23 + (9-7)^{2}*6 + (10-7)^{2}*2}{100}}\)
Standard deviation = \(\sqrt{\frac{9*1 + 4*6 + 1*26 + 0*36 + 1*23 + 4*6 + 9*2}{100}}\)
Standard deviation = \(\sqrt{\frac{9 + 24 + 26 + 0 + 23 + 24 + 18}{100}}\)
Standard deviation = \(\sqrt{\frac{124}{100}}\)
Standard deviation = \(\sqrt{1.24}\)
Standard deviation \(\approx 1.11\)
The answers are:
(a) Sample size: 100
(b) Mean: 7 cm
(c) Range: 6 cm
(d) Standard deviation: \(\approx 1.11\) cm
Key Concepts
Sample Size CalculationMean and Standard DeviationRange in Data Analysis
Sample Size Calculation
In biology, sample size is a critical factor in ensuring the reliability of research results. It refers to the number of observations or specimens in a study. When collecting data, determining the proper sample size is essential so that the findings can be generalized to the larger population. In our example of lizard length measurement, the sample size was calculated by adding up the total number of lizards at each recorded length:
\( \text{Sample size} = 1 + 6 + 26 + 36 + 23 + 6 + 2 = 100 \)
This value of 100 is crucial as it provides the research credibility and allows for meaningful statistical analysis. A larger sample size often leads to more accurate estimations, reducing the margin of error.
\( \text{Sample size} = 1 + 6 + 26 + 36 + 23 + 6 + 2 = 100 \)
This value of 100 is crucial as it provides the research credibility and allows for meaningful statistical analysis. A larger sample size often leads to more accurate estimations, reducing the margin of error.
- Consider taking a sufficiently large sample size to represent your study population well.
- Ensure your samples are collected randomly to minimize biases.
Mean and Standard Deviation
The mean and standard deviation are two fundamental concepts in statistics that describe data sets. The mean provides a measure of central tendency, indicating the average value in a data distribution. In our context, we calculate the mean length of lizards as follows:
\( \text{Mean} = \frac{\sum \text{(length * number of lizards)}}{\text{sample size}} \)
For the lizard lengths:
\( \text{Mean} = \frac{700}{100} = 7 \text{ cm} \)
On the other hand, standard deviation describes how spread out the values are around the mean. A small standard deviation signifies that the data points are close to the mean, while a large one indicates more variability.
Let's compute it for our lizards:
\( \text{Standard Deviation} = \sqrt{\frac{\sum (\text{length}_i - \text{mean})^2 \times \text{number of lizards}}{\text{sample size}}} \)
This comes out to approximately 1.11 cm in our example.
\( \text{Mean} = \frac{\sum \text{(length * number of lizards)}}{\text{sample size}} \)
For the lizard lengths:
\( \text{Mean} = \frac{700}{100} = 7 \text{ cm} \)
On the other hand, standard deviation describes how spread out the values are around the mean. A small standard deviation signifies that the data points are close to the mean, while a large one indicates more variability.
Let's compute it for our lizards:
\( \text{Standard Deviation} = \sqrt{\frac{\sum (\text{length}_i - \text{mean})^2 \times \text{number of lizards}}{\text{sample size}}} \)
This comes out to approximately 1.11 cm in our example.
- Use the mean to summarize your data and understand its general trend.
- Rely on standard deviation to gauge data variability and predictability.
Range in Data Analysis
Range is a simple yet informative statistic used in data analysis. It measures the spread or extent of a data set by calculating the difference between the maximum and minimum values. In biological studies, such as measuring the lengths of lizards, the range helps to understand the variability in species characteristics.
The calculation is straightforward:
\( \text{Range} = \text{Maximum Length} - \text{Minimum Length} \)
For the given data:
\( \text{Range} = 10 - 4 = 6 \text{ cm} \)
Even though the range provides a quick grasp of data variation, it doesn't convey information about the distribution of all values. Hence, it is often used with other statistical measures for comprehensive insights.
The calculation is straightforward:
\( \text{Range} = \text{Maximum Length} - \text{Minimum Length} \)
For the given data:
\( \text{Range} = 10 - 4 = 6 \text{ cm} \)
Even though the range provides a quick grasp of data variation, it doesn't convey information about the distribution of all values. Hence, it is often used with other statistical measures for comprehensive insights.
- Employ range to quickly understand the span of your data.
- Combine range with other metrics like mean and standard deviation for a fuller picture.
Other exercises in this chapter
Problem 10
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. $
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Find \(\bar{a}\). $$ a_{i}=2^{i}, i=1, \ldots, 5 $$
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Find \(\bar{a}\). $$ a_{i}=i / 2 \text { and } i=1, \ldots, 5 $$
View solution Problem 11
The probability expressions refer to drawing a card from a standard deck of cards. State in words the meaning of the expression and give the probability as a fr
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