Suggested languages for you:

Americas

Europe

Problem 1

Compute the (sample) variance and standard deviation of the given data sample. (You calculated the means in the Section 8.3 exercises. Round all answers to two decimal places.) -1,5,5,7,14

Expert verified

Mean = \(\frac{30}{5}\) = 6
#tag_title# Step 2: Calculate the squared deviations from the mean.
#tag_content# Subtract the mean from each data point and square the result.
(-1 - 6)^2 = 49
(5 - 6)^2 = 1
(5 - 6)^2 = 1
(7 - 6)^2 = 1
(14 - 6)^2 = 64
#tag_title# Step 3: Calculate the sample variance.
#tag_content# Add the squared deviations and divide by (n - 1), where n is the number of data points.
Variance = \(\frac{(49 + 1 + 1 + 1 + 64)}{(5 - 1)}\)
Variance = \(\frac{116}{4}\) = 29
#tag_title# Step 4: Calculate the sample standard deviation.
#tag_content# The standard deviation is the square root of the variance.
Standard Deviation = \(\sqrt{29}\) ≈ 5.39
#Completed_solution#
Sample variance = 29, and sample standard deviation ≈ 5.39 (rounded to two decimal places).

What do you think about this solution?

We value your feedback to improve our textbook solutions.

- Access over 3 million high quality textbook solutions
- Access our popular flashcard, quiz, mock-exam and notes features
- Access our smart AI features to upgrade your learning

Chapter 8

Television Ratings According to data from the Nielsen Company, there is a $1.8 \%$ chance that any television that is turned on during the time of the evening newscasts will be tuned to ABC's evening news show. \(^{65}\) Your company wishes to advertise on a local station carrying \(A B C\) that serves a community with 5,000 households that regularly watch TV during this time slot. Find the approximate probability that at least 100 households will be tuned in to the show.

Chapter 8

Are based on the following table, which shows crashworthiness ratings for several categories of motor vehicles. \({ }^{10}\) In all of these exercises, take \(X\) as the crash-test rating of a small car, \(Y\) as the crash-test rating for a small SUV, and so on, as shown in the table. $$\begin{array}{|r|c|c|c|c|c|}\hline & & {}{} {\text { Overall Frontal Crash Test Rating }} \\\\\hline &\begin{array}{c}\text { Number } \\\\\text { Tested }\end{array} & \begin{array}{c}3 \\\\(\text { Good })\end{array} & \begin{array}{c}2 \\\\\text { (Acceptable) }\end{array} & \begin{array}{c}1 \\\\\text {(Marginal) }\end{array} & \begin{array}{c}0 \\\\\text { (Poor) }\end{array} \\\\\hline \text { Small Cars, } X &16 & 1 & 11 & 2 & 2 \\\\\hline \text { Small SUVs, } Y & 10 & 1 & 4 & 4 & 1 \\\\\hline \text { MediumSUVs,} Z & 15 & 3 & 5 & 3 & 4 \\\\\hline \text { Passenger Vans, } U & 13 & 3 & 0 & 3 & 7 \\\\\hline \text { Midsize Cars, } V & 15 & 3 & 5 & 0 & 7 \\\\\hline \text { Large Cars, } W & 19 & 9 & 5 & 3&2 \\\\\hline\end{array}$$ Compare \(P(V \geq 2)\) and \(P(Z \geq 2)\). What does the result suggest?

Chapter 8

LSAT Scores LSAT test scores are normally distributed with a mean of 151 and a standard deviation of 7 . What score would place you in the top \(2 \%\) of test-takers?

Chapter 8

Distribution of Wealth If we model after-tax household income by a normal distribution, then the figures of a 1995 study imply the information in the following table, which should be used for Exercises \(49-60 .^{55}\) Assume that the distribution of incomes in each country is bell shaped and symmetric. $$ \begin{array}{|r|c|c|c|c|c|} \hline \text { Country } & \text { United States } & \text { Canada } & \text { Switzerland } & \text { Germany } & \text { Sweden } \\ \hline \begin{array}{r} \text { Mean Household } \\ \text { Income } \end{array} & \$ 38,000 & \$ 35,000 & \$ 39,000 & \$ 34,000 & \$ 32,000 \\ \hline \begin{array}{r} \text { Standard } \\ \text { Deviation } \end{array} & \$ 21,000 & \$ 17,000 & \$ 16,000 & \$ 14,000 & \$ 11,000 \\ \hline \end{array} $$ If we define a "poor" household as one whose after-tax income is at least 1.3 standard deviations below the mean, what is the household income of a poor family in Switzerland?

Chapter 8

\(\nabla\) Why is the following not a binomial random variable? Select, without replacement, five marbles from a bag containing six red marbles and two blue ones, and let \(X\) be the number of red marbles you have selected.

The first learning app that truly has everything you need to ace your exams in one place.

- Flashcards & Quizzes
- AI Study Assistant
- Smart Note-Taking
- Mock-Exams
- Study Planner