/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 14 A sample of 80 adults was taken,... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A sample of 80 adults was taken, and these adults were asked about the number of credit cards they possess. The following table gives the frequency distribution of their responses. $$ \begin{array}{lc} \hline \text { Number of Credit Cards } & \text { Number of Adults } \\ \hline 0 \text { to } 3 & 18 \\ 4 \text { to } 7 & 26 \\ 8 \text { to } 11 & 22 \\ 12 \text { to } 15 & 11 \\ 16 \text { to } 19 & 3 \\ \hline \end{array} $$ a. Find the class boundaries and class midpoints. b. Do all classes have the same width? If so, what is this width? c. Prepare the relative frequency and percentage distribution columns. d. What percentage of these adults possess 8 or more credit cards?

Short Answer

Expert verified
a) The class boundaries are -0.5 to 3.5, 3.5 to 7.5, 7.5 to 11.5, 11.5 to 15.5 and 15.5 to 19.5, and the class midpoints are 1.5, 5.5, 9.5, 13.5, and 17.5. b) Yes all classes have the same width and the width is 4. c) The relative frequency and percentages can be calculated by using their individual formulas. d) The percentage of adults possessing 8 or more credit cards can be calculated by summing the corresponding frequencies and dividing by 80, then multiplying by 100.

Step by step solution

01

Find the Class Boundaries

Class boundaries are one of the methods used to determine the classes of a frequency distribution. The class boundaries will be one unit less than the lower limit of the next class. So the class boundaries are:\[\begin{array}{lc}\hline \text { Number of Credit Cards } & \text { Class Boundaries } \\hline -0.5 \text { to } 3.5 & 18 \3.5 \text { to } 7.5 & 26 \7.5 \text { to } 11.5 & 22 \11.5 \text { to } 15.5 & 11 \15.5 \text { to } 19.5 & 3 \\hline\end{array}\]
02

Find the Class Midpoints

The class midpoint is the middle value of each class. To get this value, add the upper and lower boundaries of each class and divide by 2. This means \( (lower \ boundary+upper \ boundary)/2 \). Therefore, the class midpoints are : \[ 1.5, 5.5, 9.5, 13.5, 17.5 \]
03

Determine the Class Width

To find the class width, subtract the lower class boundary from the upper class boundary. By checking here, we can see that each class upper boundary minus the class lower boundary is equivalent to 4. Therefore, all the classes have the same width, which is 4.
04

Prepare the Relative Frequency and Percentage Distribution Columns

To calculate the relative frequency of a class, use the formula \(relative \ frequency = frequency/total \ sample \ size \). Ensure that the sum of all relative frequencies equals 1. The percentage distribution can be found by multiplying the relative frequency by 100. Hence, calculate these values for each class accordingly.
05

Calculate Percentage of Adults with 8 or More Credit Cards

To find the percentage of adults with 8 or more credit cards, add the frequencies for the categories 8 to 11, 12 to 15, 16 to 19 and divide by total number of adults then multiply by 100 to convert it to a percentage.

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 91Ó°ÊÓ!

Key Concepts

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

Class Boundaries
Class boundaries help us smooth out the edges between different classes in a frequency distribution. They serve to separate one group from the next by slightly extending each interval. The boundaries are calculated by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit of each class. This method is useful in avoiding overlaps and ensuring a clear distinction between classes.

For example, the class "0 to 3" becomes "-0.5 to 3.5." This shift is small but significant, as it helps in standardizing the gaps between classes and making frequency distribution more precise.
  • Calculating boundaries prevents gaps and overlaps.
  • Boundaries provide a smoother representation of data.
Class Midpoints
Class midpoints give us the central value of each class interval, representing the average value of that class. To find the midpoint, add the lower and upper class boundaries and divide the sum by 2. This simple calculation gives a more accurate representation of data within each class.

For the boundary of '-0.5 to 3.5,' the class midpoint is calculated as \( ( -0.5 + 3.5 ) / 2 = 1.5 \). Performing this calculation for each class will yield midpoints such as 1.5, 5.5, 9.5, and so on.
  • Midpoints help in identifying the central tendency.
  • They are useful for further statistical analysis.
Relative Frequency
Relative frequency portrays how often a particular class occurs as a proportion of the total dataset. It's a crucial element for comparing different intervals effectively. You calculate it by dividing the frequency of each class by the total number of observations:
\( \, \frac{\text{frequency of a class}}{\text{total sample size}} \, \).

Ensuring that all relative frequencies sum up to 1 is essential for consistency. This approach allows easy comparison across classes by using a unified scale.
  • Expresses frequency in relation to the total number.
  • Helps in understanding the distribution patterns within data.
Percentage Distribution
Percentage distribution converts relative frequencies into an easier-to-understand scale of 100. This makes the data more accessible for interpretation. To find the percentage distribution, multiply each relative frequency by 100.

This conversion allows one to easily visualize how much each class contributes to the entire dataset. For example, a class with a relative frequency of 0.3 would make up 30% of the total data when expressed in percentage terms.
  • Makes frequency distributions more intuitive.
  • Allows for comparisons over different datasets.

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

Why do we need to group data in the form of a frequency table? Explain briefly.

According to a survey by the U.S. Public Interest Research Group, about \(79 \%\) of credit reports contain errors. Suppose in a random sample of 25 credit reports, the number of errors found are as listed below. $$ \begin{array}{llllllllllll} 1 & 0 & 2 & 3 & 0 & 1 & 0 & 5 & 4 & 1 & 0 & 2 \\ 4 & 1 & 2 & 2 & 0 & 3 & 1 & 0 & 0 & 1 & 2 & 3 \end{array} $$ a. Prepare a frequency distribution table for these data using single-valued classes. b. Calculate the relative frequencies and percentages for all classes. c. How many of these reports contained two or more errors? d. Draw a bar graph for the frequency distribution of part a.

Statisticians often need to know the shape of a population to make inferences. Suppose that you are asked to specify the shape of the population of weights of all college students. a. Sketch a graph of what you think the weights of all college students would look like. b. The following data give the weights (in pounds) of a random sample of 44 college students (F and M indicate female and male, respectively). $$ \begin{array}{llllllll} 123 \mathrm{~F} & 195 \mathrm{M} & 138 \mathrm{M} & 115 \mathrm{~F} & 179 \mathrm{M} & 119 \mathrm{~F} & 148 \mathrm{~F} & 147 \mathrm{~F} \\ 180 \mathrm{M} & 146 \mathrm{~F} & 179 \mathrm{M} & 189 \mathrm{M} & 175 \mathrm{M} & 108 \mathrm{~F} & 193 \mathrm{M} & 114 \mathrm{~F} \\ 179 \mathrm{M} & 147 \mathrm{M} & 108 \mathrm{~F} & 128 \mathrm{~F} & 164 \mathrm{~F} & 174 \mathrm{M} & 128 \mathrm{~F} & 159 \mathrm{M} \\ 193 \mathrm{M} & 204 \mathrm{M} & 125 \mathrm{~F} & 133 \mathrm{~F} & 115 \mathrm{~F} & 168 \mathrm{M} & 123 \mathrm{~F} & 183 \mathrm{M} \\ 116 \mathrm{~F} & 182 \mathrm{M} & 174 \mathrm{M} & 102 \mathrm{~F} & 123 \mathrm{~F} & 99 \mathrm{~F} & 161 \mathrm{M} & 162 \mathrm{M} \\ 155 \mathrm{~F} & 202 \mathrm{M} & 110 \mathrm{~F} & 132 \mathrm{M} & & & & \end{array} $$ i. Construct a stem-and-leaf display for these data. ii. Can you explain why these data appear the way they do? c. Now sketch a new graph of what you think the weights of all college students look like. Is this similar to your sketch in part a?

The following data give the amounts (in dollars) spent on refreshments by 30 spectators randomly selected from those who patronized the concession stands at a recent Major League Baseball game. $$ \begin{array}{rrrrrrrr} 4.95 & 27.99 & 8.00 & 5.80 & 4.50 & 2.99 & 4.85 & 6.00 \\ 9.00 & 15.75 & 9.50 & 3.05 & 5.65 & 21.00 & 16.60 & 18.00 \\ 21.77 & 12.35 & 7.75 & 10.45 & 3.85 & 28.45 & 8.35 & 17.70 \\ 19.50 & 11.65 & 11.45 & 3.00 & 6.55 & 16.50 & & \end{array} $$ a. Construct a frequency distribution table using the less-than method to write classes. Take $$\$ 0$$ as the lower boundary of the first class and $$\$ 6$$ as the width of each class. b. Calculate the relative frequencies and percentages for all classes. c. Draw a histogram for the frequency distribution.

Table \(2.18\) contains the differences in the obesity rates (called rate change in the table) for the years between 2007 and 1997 for each of the 50 states and the District of Columbia. The obesity rate is the percentage of people having a body mass index (BMI) of 30 or higher. Figure \(2.26\) contains a dotplot of these data. a. Analyze the dotplot carefully. What value would you provide if asked to report a "typical" obesity rate change? Why did you choose this value? $$ \begin{array}{lclclc} \hline \text { State } & \begin{array}{c} \text { Rate } \\ \text { Change } \end{array} & \text { State } & \begin{array}{c} \text { Rate } \\ \text { Change } \end{array} & \text { State } & \begin{array}{c} \text { Rate } \\ \text { Change } \end{array} \\ \hline \text { AL } & 12.1 & \text { KY } & 5.6 & \text { ND } & 9.5 \\ \text { AK } & 7.8 & \text { LA } & 10.2 & \text { OH } & 9.8 \\ \text { AZ } & 13 & \text { ME } & 8.6 & \text { OK } & 13 \\ \text { AR } & 10.6 & \text { MD } & 7.9 & \text { OR } & 6.1 \\ \text { CA } & 6.6 & \text { MA } & 6.5 & \text { PA } & 9.6 \\ \text { CO } & 6.9 & \text { MI } & 8.4 & \text { RI } & 7.6 \\ \text { CT } & 6.5 & \text { MN } & 9.1 & \text { SC } & 11.5 \\ \text { DE } & 8.6 & \text { MS } & 10 & \text { SD } & 9.2 \\ \text { D.C. } & 7.3 & \text { MO } & 8.4 & \text { TN } & 12.4 \\ \text { FL } & 7.5 & \text { MT } & 7.2 & \text { TX } & 9.4 \\ \text { GA } & 13.8 & \text { NE } & 9 & \text { UT } & 6.6 \\ \text { HI } & 7.8 & \text { NV } & 10 & \text { VT } & 5.4 \\ \text { ID } & 8.2 & \text { NH } & 10.2 & \text { VA } & 7.9 \\ \text { IL } & 7.8 & \text { NJ } & 7.5 & \text { WA } & 10.1 \\ \text { IN } & 5.6 & \text { NM } & 9.1 & \text { WV } & 8.9 \\ \text { IA } & 7.5 & \text { NY } & 9 & \text { WI } & 8.1 \\ \text { KS } & 12.2 & \text { NC } & 9.7 & \text { WY } & 8.7 \\ \hline \end{array} $$ b. What number do you feel most accurately represents the number of outliers in this data set: 0 , \(1,3,4,6,9\), or \(10 ?\) Explain your reasoning, including the identification of the observations, if any, that you feel are outliers. c. Would you classify this distribution as being skewed to the left, skewed to the right, or approximately symmetric? Explain. d. The largest increase in the obesity rate during this period took place in Georgia (13.8), whereas the smallest increase took place in Vermont (5.4). Explain why this information should not lead you to conclude that Georgia had the highest obesity rate in 2007 and that Vermont had the lowest obesity rate in 2007 . (Note: The highest and lowest obesity rates in 2007 were in Mississippi and Colorado, respectively.)

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.