
Concept explainers
a)
To draw the model which shows 68-95-99.7
a)

Explanation of Solution
Given:
Following is the model which shows 68-95-99.7:
b)
To find the interval in which central 95% of IQ scores to be found.
b)

Answer to Problem 26E
The interval in which central 95% of IQ scores to be found between 68 and 132
Explanation of Solution
Given:
According to rule of 68-95-99.7, the 95% of the values are within 2 standard deviation of the
Therefore,
Hence, interval in which central 95% of IQ scores to be found between 68 and 132
c)
To find the percent of people having IQ score above 116.
c)

Answer to Problem 26E
The 16% percent of people having IQ score above 116.
Explanation of Solution
Given:
According to rule of 68-95-99.7,
That means, 116 is one standard deviation above the mean. We know that 68% of the IQ scores are within 1 standard deviation of the mean. That is, 100-68 = 32% of the IQ scores are more than 1 standard deviation from the mean.
We know, the normal curve is symmetric about the mean then 16% is more than 1 standard deviation below the mean and 16% is more than 1 standard deviation above the mean.
Hence, approximately 16% of all people should have an IQ score above 116.
d)
To find the percent of people having IQ score between 68 and 84
d)

Answer to Problem 26E
The 13.5% percent of people having IQ score between 68 and 84.
Explanation of Solution
Given:
According to rule of 68-95-99.7,
That means, 116 is one standard deviation above the mean.
That means, 68 is two standard deviation below the mean.
Therefore, 95%-68% = 27% which is more than 1 standard deviation and less than 2 standard deviation of the mean. We know, the normal curve is symmetric about the mean then 13.5% is more than 1 standard deviation but less than 2 standard deviation below the mean and 13.5% is more than 1 standard deviation but less than 2 standard deviation above the mean. So, we can say, 13.5% of people having IQ score between 68 and 84.
e)
To find the percent of people having IQ score above 132
e)

Answer to Problem 26E
The 13.5% percent of people having IQ score above 132
Explanation of Solution
Given:
According to rule, 95% of the data lies between 2 standard deviations from the mean. Since, total data is 100%, so 5% of the data is the more than 2 standard deviations from the mean. Therefore, as per symmetry, 2.5% is more than 2 standard deviations below the mean and 2.5% is more than 2 standard deviations above the mean.
That means, 132 is two standard deviation above the mean.
Therefore, approximately 2.5% of people having IQ score above 132
Chapter 6 Solutions
Stats: Modeling the World Nasta Edition Grades 9-12
Additional Math Textbook Solutions
Basic Business Statistics, Student Value Edition
Thinking Mathematically (6th Edition)
Elementary Statistics: Picturing the World (7th Edition)
A First Course in Probability (10th Edition)
Elementary Statistics (13th Edition)
- Why the correct answer is letter A? Students in an online course are each randomly assigned to receive either standard practice exercises or adaptivepractice exercises. For the adaptive practice exercises, the next question asked is determined by whether the studentgot the previous question correct. The teacher of the course wants to determine whether there is a differencebetween the two practice exercise types by comparing the proportion of students who pass the course from eachgroup. The teacher plans to test the null hypothesis that versus the alternative hypothesis , whererepresents the proportion of students who would pass the course using standard practice exercises andrepresents the proportion of students who would pass the course using adaptive practice exercises.The teacher knows that the percent confidence interval for the difference in proportion of students passing thecourse for the two practice exercise types (standard minus adaptive) is and the percent…arrow_forwardCarpetland salespersons average $8,000 per week in sales. Steve Contois, the firm's vice president, proposes a compensation plan with new selling incentives. Steve hopes that the results of a trial selling period will enable him to conclude that the compensation plan increases the average sales per salesperson. a. Develop the appropriate null and alternative hypotheses.H 0: H a:arrow_forwardتوليد تمرين شامل حول الانحدار الخطي المتعدد بطريقة المربعات الصغرىarrow_forward
- The U.S. Postal Service will ship a Priority Mail® Large Flat Rate Box (12" 3 12" 3 5½") any where in the United States for a fixed price, regardless of weight. The weights (ounces) of 20 ran domly chosen boxes are shown below. (a) Make a stem-and-leaf diagram. (b) Make a histogram. (c) Describe the shape of the distribution. Weights 72 86 28 67 64 65 45 86 31 32 39 92 90 91 84 62 80 74 63 86arrow_forward(a) What is a bimodal histogram? (b) Explain the difference between left-skewed, symmetric, and right-skewed histograms. (c) What is an outlierarrow_forward(a) Test the hypothesis. Consider the hypothesis test Ho = : against H₁o < 02. Suppose that the sample sizes aren₁ = 7 and n₂ = 13 and that $² = 22.4 and $22 = 28.2. Use α = 0.05. Ho is not ✓ rejected. 9-9 IV (b) Find a 95% confidence interval on of 102. Round your answer to two decimal places (e.g. 98.76).arrow_forward
- Let us suppose we have some article reported on a study of potential sources of injury to equine veterinarians conducted at a university veterinary hospital. Forces on the hand were measured for several common activities that veterinarians engage in when examining or treating horses. We will consider the forces on the hands for two tasks, lifting and using ultrasound. Assume that both sample sizes are 6, the sample mean force for lifting was 6.2 pounds with standard deviation 1.5 pounds, and the sample mean force for using ultrasound was 6.4 pounds with standard deviation 0.3 pounds. Assume that the standard deviations are known. Suppose that you wanted to detect a true difference in mean force of 0.25 pounds on the hands for these two activities. Under the null hypothesis, 40 = 0. What level of type II error would you recommend here? Round your answer to four decimal places (e.g. 98.7654). Use a = 0.05. β = i What sample size would be required? Assume the sample sizes are to be equal.…arrow_forward= Consider the hypothesis test Ho: μ₁ = μ₂ against H₁ μ₁ μ2. Suppose that sample sizes are n₁ = 15 and n₂ = 15, that x1 = 4.7 and X2 = 7.8 and that s² = 4 and s² = 6.26. Assume that o and that the data are drawn from normal distributions. Use απ 0.05. (a) Test the hypothesis and find the P-value. (b) What is the power of the test in part (a) for a true difference in means of 3? (c) Assuming equal sample sizes, what sample size should be used to obtain ẞ = 0.05 if the true difference in means is - 2? Assume that α = 0.05. (a) The null hypothesis is 98.7654). rejected. The P-value is 0.0008 (b) The power is 0.94 . Round your answer to four decimal places (e.g. Round your answer to two decimal places (e.g. 98.76). (c) n₁ = n2 = 1 . Round your answer to the nearest integer.arrow_forwardConsider the hypothesis test Ho: = 622 against H₁: 6 > 62. Suppose that the sample sizes are n₁ = 20 and n₂ = 8, and that = 4.5; s=2.3. Use a = 0.01. (a) Test the hypothesis. Round your answers to two decimal places (e.g. 98.76). The test statistic is fo = i The critical value is f = Conclusion: i the null hypothesis at a = 0.01. (b) Construct the confidence interval on 02/022 which can be used to test the hypothesis: (Round your answer to two decimal places (e.g. 98.76).) iarrow_forward
- 2011 listing by carmax of the ages and prices of various corollas in a ceratin regionarrow_forwardس 11/ أ . اذا كانت 1 + x) = 2 x 3 + 2 x 2 + x) هي متعددة حدود محسوبة باستخدام طريقة الفروقات المنتهية (finite differences) من جدول البيانات التالي للدالة (f(x . احسب قيمة . ( 2 درجة ) xi k=0 k=1 k=2 k=3 0 3 1 2 2 2 3 αarrow_forward1. Differentiate between discrete and continuous random variables, providing examples for each type. 2. Consider a discrete random variable representing the number of patients visiting a clinic each day. The probabilities for the number of visits are as follows: 0 visits: P(0) = 0.2 1 visit: P(1) = 0.3 2 visits: P(2) = 0.5 Using this information, calculate the expected value (mean) of the number of patient visits per day. Show all your workings clearly. Rubric to follow Definition of Random variables ( clearly and accurately differentiate between discrete and continuous random variables with appropriate examples for each) Identification of discrete random variable (correctly identifies "number of patient visits" as a discrete random variable and explains reasoning clearly.) Calculation of probabilities (uses the probabilities correctly in the calculation, showing all steps clearly and logically) Expected value calculation (calculate the expected value (mean)…arrow_forward
- MATLAB: An Introduction with ApplicationsStatisticsISBN:9781119256830Author:Amos GilatPublisher:John Wiley & Sons IncProbability and Statistics for Engineering and th...StatisticsISBN:9781305251809Author:Jay L. DevorePublisher:Cengage LearningStatistics for The Behavioral Sciences (MindTap C...StatisticsISBN:9781305504912Author:Frederick J Gravetter, Larry B. WallnauPublisher:Cengage Learning
- Elementary Statistics: Picturing the World (7th E...StatisticsISBN:9780134683416Author:Ron Larson, Betsy FarberPublisher:PEARSONThe Basic Practice of StatisticsStatisticsISBN:9781319042578Author:David S. Moore, William I. Notz, Michael A. FlignerPublisher:W. H. FreemanIntroduction to the Practice of StatisticsStatisticsISBN:9781319013387Author:David S. Moore, George P. McCabe, Bruce A. CraigPublisher:W. H. Freeman





