Question

Petroleum pollution in seas and oceans stimulates the growth of some types of bacteria. A count of the number of bacteria (per 100 milliliters) in 10 portions of seawater gave these readings: $$ 49,70,54,67,59,40,61,69,71,52 $$ a. Calculate the range. b. Calculate \(\bar{x}\) and \(s\). c. The range is about how many standard deviations?

Short answer

Answer: The range of the dataset is approximately 3.15 standard deviations.

Step-by-step solution

Find the Maximum and Minimum Values

The first step is to find the maximum and minimum values in the dataset. The dataset is: $$ 49, 70, 54, 67, 59, 40, 61, 69, 71, 52 $$ Sorting the dataset in ascending order, we have: $$ 40, 49, 52, 54, 59, 61, 67, 69, 70, 71 $$ The minimum value is 40 and the maximum value is 71.

Calculate the Range

The range is the difference between the maximum and minimum values. Using the values found in Step 1, we can calculate the range as follows: $$ Range = Maximum - Minimum = 71 - 40 = 31 $$ The range of the dataset is 31.

Calculate the Mean

To calculate the mean (\(\bar{x}\)), we need to add up all the values in the dataset and divide the sum by the total number of values (n). In this case, n = 10. $$ \bar{x} = \frac{49 + 70 + 54 + 67 + 59 + 40 + 61 + 69 + 71 + 52}{10} = \frac{592}{10} = 59.2 $$ The mean of the dataset is 59.2.

Calculate the Standard Deviation

To calculate the standard deviation (\(s\)), we first need to find the variance. Variance is the average of the squared differences from the mean. The formula is: $$ \text{Variance} = \frac{\sum(x_i - \bar{x})^2}{n} $$ Let's do this step by step: 1. Find the difference between each value and the mean. 2. Square the differences. 3. Find the sum of the squared differences. 4. Divide the sum by the total number of values. 1. Differences: $$ (49 - 59.2), (70 - 59.2), (54 - 59.2), (67 - 59.2), (59 - 59.2), (40 - 59.2), (61 - 59.2), (69 - 59.2), (71 - 59.2), (52 - 59.2) $$ 2. Squared differences: $$ 104.04, 115.56, 27.04, 60.84, 0.04, 370.44, 3.24, 96.04, 139.24, 51.84 $$ 3. Sum of squared differences: $$ 104.04 + 115.56 + 27.04 + 60.84 + 0.04 + 370.44 + 3.24 + 96.04 + 139.24 + 51.84 = 968.32 $$ 4. Divide the sum by the total number of values: $$ \text{Variance} = \frac{968.32}{10} = 96.832 $$ To get the standard deviation, take the square root of the variance: $$ s = \sqrt{96.832} \approx 9.84 $$ The standard deviation of the dataset is approximately 9.84.

Determine How Many Standard Deviations the Range is

Now, we need to find out how many standard deviations the range is. By dividing the range by the standard deviation, we can get the answer. $$ \frac{Range}{s} = \frac{31}{9.84} \approx 3.15 $$ The range is about 3.15 standard deviations.

Key concepts

Range of a Data Set

Understanding the range of a data set is fundamental in descriptive statistics. The range gives us a quick sense of how spread out the values in our dataset are. It is simply calculated by subtracting the smallest value from the largest value. In the exercise provided, we computed the range for bacteria counts in portions of seawater and found it to be 31.

To reiterate the concept with our example, we took the highest bacteria count, 71, and subtracted the lowest count, 40, to obtain a range of 31. This single number tells us that the bacteria counts in the seawater samples can vary by as much as 31 units.

However, the range doesn't give us details about the distribution between these two points. It's just a starting point for understanding variability, and it's most useful when comparing consistency between different data sets or samples.

Mean and Standard Deviation

When we move beyond the range, we delve into the mean and standard deviation, which tell us more about the overall distribution of our data. The mean, denoted as \(\bar{x}\), is what most people refer to as the average. To find it, we sum all the values and divide by the count of the values. For the seawater bacteria counts, we calculated the mean to be 59.2.

The standard deviation (\(s\)) builds on the concept of the mean and is a measure of how much each value in the data set tends to deviate from the mean. It is a more reliable metric of variability than the range because it takes into account how each data point contributes to the overall spread. We calculated the standard deviation to be approximately 9.84. This number allows us to understand that, on average, the bacteria counts deviate from the mean by about 9.84 units.

A small standard deviation relative to the mean indicates that the data points tend to be close to the mean, while a large standard deviation indicates that the data is spread out over a wider range of values.

Variance in Statistics

To talk about variance in statistics is to discuss the foundation upon which standard deviation is built. Variance measures the average degree to which each point differs from the mean - in other words, the spread of the data points. We calculate it by averaging the squared differences between each data point and the mean.

In our exercise, after squaring these differences and averaging them, we found the variance to be 96.832. Variance is the squared value of the standard deviation and therefore gives weight to outliers; this is because larger differences from the mean are squared, thus contributing more significantly to the overall value of the variance.

Understanding variance is key to many statistical concepts and tests because it quantifies the level of volatility or dispersion present in the data. In predictive models, for instance, a lower variance is often associated with more reliable predictions.