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Tuna Fish An article in Consumer Reports gives the price-an estimated average for a 6 -ounce can or a 7.06 -ounce pouch \(-\) for 14 dif. ferent brands of water-packed light tuna, based on prices paid nationally in supermarkets:= \(\begin{array}{llll}99 & 1.92 & 1.73 & 85\end{array}\) 1.4 \(\begin{array}{lllllll}1.12 & .63 & .67 & 69 & 60 & 60 & .66\end{array}\) a. Find the average price for the 14 different brands of tunal. b. Find the median price for the 14 different brands of tuna. c. Based on your findings in parts a and \(b\), do you think that the distribution of prices is skewed'? Explain.

Short Answer

Expert verified
Explain your answer. Answer: The distribution of the 14 different tuna brands' prices is not significantly skewed. It's approximately symmetric, as the average and median prices are very close to each other ($0.91 and $0.92, respectively).

Step by step solution

01

Arrange the data in ascending order

First, let's arrange the prices in ascending order: \(0.60, 0.60, 0.63, 0.66, 0.67, 0.69, 0.85, 0.99, 1.12, 1.40,1.73, 1.92\)
02

Calculate the average price

To find the average price, we will add up all the prices and divide by the total number of prices (14). Average \(= \frac{0.60 + 0.60 + 0.63 + 0.66 + 0.67 + 0.69 + 0.85 + 0.99 + 1.12 + 1.40 + 1.73 + 1.92}{14}\) Average \(= \frac{12.76}{14}\) Average \(= 0.91\) So, the average price of the 14 different tuna brands is $0.91.
03

Calculate the median price

To find the median price, we need to find the middle value of the sorted list. Since we have an even number of values (14), we will take the average of the two middle values (7th and 8th values). Median \(= \frac{0.85 + 0.99}{2}\) Median \(= \frac{1.84}{2}\) Median \(= 0.92\) So, the median price of the 14 different tuna brands is $0.92.
04

Analyze distribution skewness

To determine if the distribution of prices is skewed, we will compare the average and median prices. Average price: $0.91 Median price: \(0.92\) The average and median prices are almost equal, with the average being slightly lower than the median. This suggests that the distribution is not significantly skewed, and it's approximately symmetric. In conclusion, based on the findings in parts a and b, the distribution of prices is not significantly skewed. It's approximately symmetric as the average and median prices are very close to each other.

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Key Concepts

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

Average Calculation
Understanding how to calculate the average of a set of numbers is fundamental in descriptive statistics. The average, or mean, provides a summary measure of a data set, representing the central tendency.
To find the average, we follow these simple steps:
  • Add up all the numbers in the data set.
  • Divide the sum by the total number of data points.
Using the example given in the original exercise, the prices of 14 different brands of tuna were totaled, resulting in a sum of 12.76. The next step was to divide this total by the number of prices, which is 14, giving us an average price of approximately $0.91 per can.
This calculation implies that if all prices were identical, each would be $0.91 per can. This average gives us a quick, at-a-glance idea of what the typical price in this set is.
Median Calculation
The median is another key measure of central tendency in descriptive statistics. It represents the middle value in an ordered data set.
To find the median:
  • First, arrange all numbers in ascending order.
  • If the data set has an odd number of elements, the median is the middle one.
  • If even, it's the average of the two central numbers.
In our example, after arranging the tuna prices in order, we found there are 14 numbers. This even count means the median is the average of the 7th and 8th prices: \(0.85 and \)0.99. Calculating \[\frac{0.85 + 0.99}{2} = 0.92\]provides a median of $0.92.
The median offers another way to understand the data set's typical value, helping highlight the exact middle of the data distribution, minimizing the influence of outliers.
Symmetric Distribution
Understanding whether a distribution is symmetric or skewed gives valuable insights into the data's nature. A symmetric distribution means the data is evenly distributed around the central point, with the mean and median being nearly equal.
In the tuna price example, we see that the average price is $0.91, while the median is $0.92. This small difference suggests a symmetric distribution because:
  • A perfectly symmetric distribution usually has the mean equal to the median.
  • Small deviations, like those seen here, indicate the data does not significantly favor the high or low end.
  • There is approximately equal spread of data around the central value.
Such symmetry in distributions can make statistical analysis simpler and interpretations more straightforward, as it implies balanced data without extreme skewness.

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Most popular questions from this chapter

The number of raisins in each 14 miniboxes (1/2-ounce size) was counted for a generic brand and for Sunmaid brand raisins. The two data sets are shown here: $$ \begin{array}{llll|llll} \multicolumn{3}{l|} {\text { Generic Brand }} & \multicolumn{3}{|c} {\text { Sunmaid }} \\ \hline 25 & 26 & 25 & 28 & 25 & 29 & 24 & 24 \\ 26 & 28 & 28 & 27 & 28 & 24 & 28 & 22 \\ 26 & 27 & 24 & 25 & 25 & 28 & 30 & 27 \\ 26 & 26 & & & 28 & 24 & & \end{array} $$ a. What are the mean and standard deviation for the generic brand? b. What are the mean and standard deviation for the Sunmaid brand? c. Compare the centers and variabilities of the two brands using the results of parts a and \(b\).

You are given \(n=8\) measurements: 3,1,5,6 4,4,3,5 a. Calculate the range. b. Calculate the sample mean. c. Calculate the sample variance and standard deviation. d. Compare the range and the standard deviation. The range is approximately how many standard deviations?

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.

Here are the ages of 50 pennies from Exercise 1.45 and data set. The data have been sorted from smallest to largest. \(\begin{array}{llllllllll}0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\end{array}\) \(\begin{array}{rrrrrrrrrr}0 & 0 & 1 & 1 & 1 & 1 & 1 & 1 & 2 & 2 \\ 2 & 3 & 3 & 3 & 4 & 4 & 5 & 5 & 5 & 5 \\ 6 & 8 & 9 & 9 & 10 & 16 & 17 & 17 & 19 & 19 \\ 19 & 20 & 20 & 21 & 22 & 23 & 25 & 25 & 28 & 36\end{array}\) a. What is the average age of the pennies? b. What is the median age of the pennies? c. Based on the results of parts a and \(b\), how would you describe the age distribution of these 50 pennies? d. Construct a box plot for the data set. Are there any outliers? Does the box plot confirm your description of the distribution's shape?

Mercury Concentration in Dolphins Environmental scientists are increasingly concerned with the accumulation of toxic elements in marine mammals and the transfer of such elements to the animals' offspring. The striped dolphin (Stenella coeruleoalba), considered to be a top predator in the marine food chain, was the subject of one such study. The mercury concentrations (micrograms/gram) in the livers of 28 male striped dolphins were as follows: \(\begin{array}{llll}1.70 & 183.00 & 221.00 & 286.00 \\ 1.72 & 168.00 & 406.00 & 315.00\end{array}\) \(\begin{array}{rlll}85.40 & 481.00 & 445.00 & 314.00 \\\ 118.00 & 485.00 & 278.00 & 318.00\end{array}\) a. Calculate the five-number summary for the data. b. Construct a box plot for the data. c. Are there any outliers? d. If you knew that the first four dolphins were all less than 3 years old, while all the others were more than 8 years old, would this information help explain the difference in the magnitude of those four observations? Explain.

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