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Chapter 3: Problem 22

Although bats are not known for their eyesight, they are able to locate prey (mainly insects) by emitting high-pitched sounds and listening for echoes. A paper appearing in Animal Behaviour ("The Echolocation of Flying Insects by Bats" [1960]: 141-154) gave the following distances (in centimeters) at which a bat first detected a nearby insect: $$ \begin{array}{lllllllllll} 62 & 23 & 27 & 56 & 52 & 34 & 42 & 40 & 68 & 45 & 83 \end{array} $$ a. Calculate and interpret the mean distance at which the bat first detects an insect. b. Calculate the sample variance and standard deviation for this data set. Interpret these values.

Short Answer

Expert verified
The mean distance at which the bat first detects an insect is approximately \(48.36\) centimeters. The sample variance for this dataset is \(529.02\) centimeters虏, and the standard deviation is approximately \(23.01\) centimeters. This means that the distances at which bats detect insects tend to deviate about \(23.01\) centimeters from the mean distance.

Step by step solution

01

Organize the data

First, let's organize the data into a more manageable format. We have the following distances in centimeters: \( 62, 23, 27, 56, 52, 34, 42, 40, 68, 45, 83 \)
02

Calculate the mean distance

The mean distance is the sum of the distances divided by the total number of observations. To calculate the mean, we'll follow these steps: 1. Find the sum of all distances. 2. Divide this sum by the number of observations. Mean \( = \frac{\sum{ Distances }}{ Number\ of\ observations} \)
03

Find the sum of the distances

Adding the distances together, we get: \( 62+23+27+56+52+34+42+40+68+45+83 = 532 \)
04

Calculate the mean

Now, we have the sum of the distances and the number of observations (11). We can calculate the mean by dividing the sum by the number of observations: Mean \( = \frac{532}{11} = 48.36 \) The mean distance at which the bat first detects an insect is approximately \(48.36\) centimeters.
05

Calculate the sample variance

To calculate the sample variance, we'll first find the difference between each observation and the mean, square each difference, and then sum these squared differences. Finally, we'll divide this sum by the number of observations minus 1. Sample Variance \( = \frac{\sum{ (distance\ -\ mean)^2 }}{ Number\ of\ observations -1 }\)
06

Find the squared differences

Calculate the squared differences between each observation and the mean: \( (62-48.36)^2=187.21, (23-48.36)^2=643.61, ..., (83-48.36)^2=1199.61 \)
07

Find the sum of the squared differences

Next, find the sum of these squared differences: \( 187.21 + 643.61 + ... + 1199.61 = 5290.18 \)
08

Calculate the sample variance

Now, we can divide the sum of the squared differences by the number of observations minus 1: Sample Variance \( = \frac{5290.18}{11-1} = 529.02 \) The sample variance for this dataset is approximately \(529.02\).
09

Calculate the standard deviation

Finally, we can find the standard deviation by taking the square root of the sample variance: Standard Deviation \( = \sqrt{529.02} = 23.01 \) The standard deviation for this dataset is approximately \(23.01\) centimeters. Interpretation of the values: The mean distance at which the bat detected the insect is about \(48.36\) centimeters. With a standard deviation of \(23.01\) centimeters, we can say that the distances at which bats detect insects tend to deviate approximately \(23.01\) centimeters from the mean distance. The sample variance, which is the square of the standard deviation, indicates the spread of data, with a value of \(529.02\) centimeters虏.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Mean
In the context of descriptive statistics, the mean is a measure of central tendency. It sums up a set of values and then divides by the number of values to find a central point.
In this exercise, we calculated the mean distance at which the bat detected insects. To compute this, we first summed all the distances:
  • Sum of distances: 62 + 23 + 27 + 56 + 52 + 34 + 42 + 40 + 68 + 45 + 83 = 532
The next step was to divide this sum by the number of distance readings, which was 11.
  • Mean distance = \( \frac{532}{11} \approx 48.36 \) cm
The mean distance of approximately 48.36 cm indicates a central point around which the bat typically detects insects. This average helps to understand the overall detection capability in terms of distance.
Variance
Variance offers insight into the spread or dispersion of a dataset around the mean. It is calculated by finding the average of the squared differences from the Mean.
To find the variance of the bat detection distances, we calculated the differences between each distance and the mean, squaring each difference:
  • Example calculation: \( (62 - 48.36)^2 = 187.21 \)
Then, we summed these squared differences:
  • Sum of squared differences: 5290.18
Finally, we divided by the number of observations minus one (which is 10 for this dataset) to find the sample variance:
  • Sample Variance = \( \frac{5290.18}{10} \approx 529.02 \)
The variance of 529.02 tells us how much the distances vary from the mean on average. A larger variance indicates more spread out data.
Standard Deviation
The standard deviation is directly derived from the variance and helps quantify how much the individual data points deviate from the mean.
Since variance measures squared differences, its unit tends to be larger and squared. By taking the square root of variance, we bring it back to the original unit of measurement:
  • Standard Deviation = \( \sqrt{529.02} \approx 23.01 \) cm
With a standard deviation of approximately 23.01 cm, we infer that the bat detection distances typically vary about 23.01 cm from the mean distance.
This standard deviation value, along with the mean, provides a picture of the detection distance distribution, giving a clearer perspective of typical distance variations observed by bats.

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Most popular questions from this chapter

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