Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Petroleum pollution in seas and oceans stimulates the growth of some types of bacteria. A count of petroleumlytic micro-organisms (bacteria per 100 milliliters) in 10 portions of seawater gave these readings: \(\begin{array}{llll}49, & 70, & 54, & 67 .\end{array}\) \(\begin{array}{llllll}59, & 40, & 61, & 69, & 71, & 52\end{array}\) a. Guess the value for \(s\) using the range approximation. b. Calculate \(\bar{x}\) and \(s\) and compare with the range approximation of part a. c. Construct a box plot for the data and use it to describe the data distribution.

Short Answer

Expert verified
Answer: Using the range approximation, the estimated standard deviation is 7.75. The actual standard deviation is 8.16, which is slightly higher than the estimated value.

Step by step solution

01

Organize data

First, organize the given data in ascending order. 40, 49, 52, 54, 59, 61, 67, 69, 70, 71
02

Find the range

Find the range of the data, which is the difference between the highest and lowest values. Range = Highest value - Lowest value Range = 71 - 40 Range = 31
03

Range approximation for standard deviation

Estimate the standard deviation (s) using the range approximation. Divide the range by 4. s ≈ Range / 4 s ≈ 31 / 4 s ≈ 7.75
04

Calculate the mean

Calculate the mean (\(\bar{x}\)) of the data by adding all the values and dividing by the total number (10) of data points. \(\bar{x} = \frac{40 + 49 + 52 + 54 + 59 + 61 + 67 + 69 + 70 + 71}{10}\) \(\bar{x} = \frac{592}{10}\) \(\bar{x} = 59.2\)
05

Calculate the actual standard deviation

Calculate the actual standard deviation (s) using the formula: \(s = \sqrt{\frac{\sum(x_i - \bar{x})^2}{n - 1}}\), where \(x_i\) are the individual data points, \(\bar{x}\) is the mean, and n is the total number of data points. \(s = \sqrt{\frac{(40-59.2)^2 + (49-59.2)^2 + (52-59.2)^2 + (54-59.2)^2 + (59-59.2)^2 + (61-59.2)^2 + (67-59.2)^2 + (69-59.2)^2 + (70-59.2)^2+ (71-59.2)^2}{10-1}}\) \(s = \sqrt{\frac{538.8}{9}}\) \(s ≈ 8.16\) The actual standard deviation (s) is 8.16, which is slightly higher than our range approximation of 7.75.
06

Construct a box plot

In order to construct a box plot, we need to find the following values: 1. Lower Quartile (Q1): The median of the lower half of the data. Data for lower half: 40, 49, 52, 54, 59 Q1 = 52 (the middle value in the lower half) 2. Median (Q2): The middle value of the entire dataset. Data: 40, 49, 52, 54, 59, 61, 67, 69, 70, 71 Q2 = 60 (the average of 59 and 61, the middle values) 3. Upper Quartile (Q3): The median of the upper half of the data. Data for upper half: 61, 67, 69, 70, 71 Q3 = 69 (the middle value in the upper half) 4. Minimum and Maximum values are the lowest and highest values in the dataset, respectively. Minimum = 40 Maximum = 71 Now we can construct a box plot using these values. The box plot will show that the data distribution is slightly skewed to the left, with a larger spread in the lower half of the data. Comparing the mean (59.2) and the median (60) values, it also indicates a slight left skew. The box plot and calculated values provide a good understanding of the given dataset's distribution and central tendency.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Standard Deviation
Standard deviation is a statistical measure that tells us how much the values in a dataset deviate from the mean or average. It provides insight into the data's variability or spread. A dataset with a high standard deviation means the data points are spread out widely around the mean, while a low standard deviation indicates that the data points are clustered closely around the mean.

To calculate the standard deviation, first find the mean (\(\bar{x}\)) of the dataset. Then, subtract the mean from each data point to get the deviation of each point from the mean. Square these deviations, sum them up, and divide by the number of data points minus one to get the variance. The standard deviation is the square root of this variance. \[s = \sqrt{\frac{\sum(x_i - \bar{x})^2}{n - 1}}\]
  • The mean is 59.2.
  • The calculated deviations are squared and summed up to give 538.8.
  • The standard deviation was found to be approximately 8.16.
Standard deviation serves as an important tool in understanding the spread and consistency of data, particularly when interpreting a dataset's overall distribution.
Mean Calculation
The mean, often referred to as the average, is a measure of central tendency and one of the most commonly used statistics in analyzing a dataset. It provides a simple summary of all the data points by averaging them over the entire dataset.

To calculate the mean for a series of numbers, sum up all the numbers and then divide by the count of the numbers. For the given dataset:
  • Data values: 40, 49, 52, 54, 59, 61, 67, 69, 70, 71
  • Mean is calculated as: \( \bar{x} = \frac{40 + 49 + 52 + 54 + 59 + 61 + 67 + 69 + 70 + 71}{10} = 59.2 \)
The mean serves as the balancing point of the data, giving an overall sense of the dataset's value. It is sensitive to all data points and can be influenced by extreme values or outliers. However, unlike the median or mode, the mean benefits from reflecting every piece of information in the dataset.
Box Plot
A box plot is a graphical representation of a dataset that effectively shows the dataset's distribution in terms of its quartiles and median. It provides a visual summary of the minimum, first quartile (Q1), median, third quartile (Q3), and maximum, which together give valuable insights into the dataset's spread and central tendency.
  • Q1: 52
  • Median (Q2): 60
  • Q3: 69
  • Minimum: 40
  • Maximum: 71
The box in the box plot captures the interquartile range (IQR), which is the middle 50% of the data. Lines ("whiskers") extend to the minimum and maximum values that aren't considered outliers.

In this particular dataset, the box plot reveals a slight left skew, indicating that there is more data spread on the lower end. By visually assessing a box plot, one can easily detect skewness, identify outliers, and compare the quartiles. It serves as a powerful tool for interpreting data distributions alongside numerical measures like mean or standard deviation.
Data Distribution
Data distribution refers to how values of a dataset are spread or dispersed. Understanding data distribution helps to identify patterns, trends, and potential anomalies within the data.

In descriptive statistics, data distribution can be described by characteristics like shape, center, and spread. The shape might be normal (bell-shaped), skewed left or skewed right, or uniform. The center is often represented by the mean or median, and the spread is detailed by measures like range, variance, or standard deviation.
  • The mean of 59.2 and the median of 60 suggest a distribution close to central tendency.
  • The actual higher standard deviation of 8.16 compared to 7.75 hints at a greater variation among data points.
  • This dataset's distribution was slightly preferred toward the lower numbers, as indicated by our box plot's skewness to the left.
Visual aids like histograms or box plots complement numerical summaries, helping to grasp the dataset's distribution effectively. When analyzing any dataset, understanding its distribution is paramount to forming accurate statistical inferences and decisions.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A company interested in lumbering rights for a certain tract of slash pine trees is told that the mean diameter of these trees is 14 inches with a standard deviation of 2.8 inches. Assume the distribution of diameters is roughly mound-shaped. a. What fraction of the trees will have diameters between 8.4 and 22.4 inches? b. What fraction of the trees will have diameters greater than 16.8 inches?

A strain of long stemmed roses has an approximate normal distribution with a mean stem length of 15 inches and standard deviation of 2.5 inches. a. If one accepts as "long-stemmed roses" only those roses with a stem length greater than 12.5 inches, what percentage of such roses would be unacceptable? b. What percentage of these roses would have a stem length between 12.5 and 20 inches?

The miles per gallon (mpg) for each of 20 medium-sized cars selected from a production line during the month of March follow. \(\begin{array}{llll}23.1 & 21.3 & 23.6 & 23.7\end{array}\) \(\begin{array}{llll}20.2 & 24.4 & 25.3 & 27.0 \\ 24.7 & 22.7 & 26.2 & 23.2\end{array}\) 25.9 \(\begin{array}{llll}24.9 & 22.2 & 22.9 & 24.6\end{array}\) a. What are the maximum and minimum miles per gallon? What is the range? b. Construct a relative frequency histogram for these data. How would you describe the shape of the distribution? c. Find the mean and the standard deviation. d. Arrange the data from smallest to largest. Find the \(z\) -scores for the largest and smallest observations. Would you consider them to be outliers? Why or why not? e. What is the median? f. Find the lower and upper quartiles.

A group of laboratory animals is infected with a particular form of bacteria, and their survival time is found to average 32 days, with a standard deviation of 36 days. a. Visualize the distribution of survival times. Do you think that the distribution is relatively mound shaped, skewed right, or skewed left? Explain. b. Within what limits would you expect at least \(3 / 4\) of the measurements to lie?

You can use the Empirical Rule to see why the distribution of survival times could not be mound shaped. a. Find the value of \(x\) that is exactly one standard deviation below the mean. b. If the distribution is in fact mound-shaped. approximately what percentage of the measurements should be less than the value of \(x\) found in part a? c. Since the variable being measured is time, is it possible to find any measurements that are more than one standard deviation below the mean? d. Use your answers to parts \(b\) and \(c\) to explain why the data distribution cannot be mound-shaped.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free