Suppose students who enter kindergarten at Stanley Elementary School take a standardized vocabulary assessment. The scores on the vocabularyassessment produce a normal density curve as shown.5475From the graph shown, give the mean and median, and then use the 68-95-99.7% (empirical) rule to estimate the standard deviation of the vocabularyassessment.For a nonstandard normal density curve, about 68% of scores are within 1within 2196The median vocabulary assessment score ispoints.standard deviations of the mean, and 99.7% are within 3Because 99.7% of values are within 3points and the mean vocabulary assessment score is 75standard deviations of the mean* standard deviations between 96 points and 54 points.standard deviation of the mean, about 95% arestandard deviations of the mean.To estimate the standard deviation, take the difference between 96 points and 54 points and divide by 6The standard deviation for the nonstandard normal density curve shown ispoints.Ithere should be about

the empirical rule predicts that in normal distributions, 68% of observations fall within the first…

Answered: Suppose students who enter kindergarten…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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n The gas mileages (in miles per gallon) for 26 cars are shown in the frequency distribution. Approximate the mean of the frequency distribution. Gas Mileage (in miles per gallon) Frequency 8 29-32 33-36 13 37-40 1 41-44 4 ... The approximate mean of the frequency distribution is (Round to one decimal place as needed.)
Determine the number of standard deviations from the mean (z-score). In the United States, the distribution of a male's height has a mean of 69 inches with a standard deviation of 3 inches. How many standard deviations from the mean is John Doe, who is 67 inches tall? Question 15 options: 0.67 standard deviations below the mean 0.33 standard deviations below the mean 2 standard deviations below the mean 1 standard deviation below the mean
Refer to the data set in the accompanying table. Assume that the paired sample data is a simple random sample and the differences have a distribution that is approximately normal. Use a significance level of 0.05 to test for a difference between the weights of discarded paper (in pounds) and weights of discarded plastic (in pounds). E Click the icon to view the data. In this example, Hg is the mean value of the differences d for the population of all pairs of data, where each individual difference d is defined as the weight of discarded paper minus the weight of discarded plastic for a household. What are the null and alternative hypotheses for the hypothesis test? O A. Ho: Ha = 0 H,: Ha #0 O B. Ho: Ha #0 H1: Hd =0 O C. Ho: Ha #0 O D. Ho: Ha = 0 H1: Hd 0
The organization of the Girl Scouts has completed its annual cookie drive. The sales are reported in the accompanying ogive. Approximate the percentage of girls who sold more than 70 boxes of cookies.
Among freshmen at a certain university, scores on the Math SAT followed the normal curve, with an average of 550 and an SD of 100. The SAT score of a student whose score is at the 99th percentile is closest to what value? Group of answer choices 780 400 760 805
Use the Histogram tool to develop a frequency distribution and histogram for the number of months as a customer of the bank in the data below. Compute the relative and cumulative relative frequencies and use a line chart to construct an ogive. Complete the frequency distribution for the months as a customer, and compute the relative and cumulative relative frequencies. (Type integers or decimals rounded to three decimal places as needed.) Bins Frequency Relative Frequency Cumulative Relative Frequency 10 enter your response here enter your response here enter your response here 20 enter your response here enter your response here enter your response here 30 enter your response here enter your response here enter your response here 40 enter your response here enter your response here enter your response here 50 enter your response here enter your response here enter your…
1:15 (a) Which variable is the explanatory variable? OA. One of the biggest factors in determining the value of a home is the square footage. The accompanying data represent the square footage and selling price (in thousands of dollars) for a random sample of homes for sale in a certain region. Complete parts (a) through (h) below. Click the icon to view the housing data. Square Footage Selling Price (b) Draw a scatter diagram of the data. Choose the correct scatter diagram below. Ay 4500- 000 1000 mylab.pearson.com/ + ® : 3 4500 Data Table Q Square Footage, x 2148 3231 1177 1988 3064 2735 4146 2215 2625 1629 1773 3887 Print ||| OB. r= 0.899 (Round to three decimal places as needed.) = AX 4500- 1000+++ 150 750 Q (c) Determine the linear correlation coefficient between square footage and asking price. Done Ay 750- Vo) 1 LTE2 150- O 1000 Selling Price ($000s), y D 371.1 382.2 197.6 339.9 614.9 364.7 632.7 377.7 427.5 285.9 268.9 701.4 4500 .... .... Q X D. Ay 750- 41% 150+ 1000 4500
Find the mean of the data summarized in the given frequency distribution. Compare the computed mean to the actual mean of 51.9 miles per hour. Speed (miles per hour) Frequency 42 - 45 46-49 50 - 53 54 -57 58 -61 25 15 The mean of the frequency distribution is miles per hour. (Round to the nearest tenth as needed.)
Find the mean of the data summarized in the given frequency distribution. Compare the computed mean to the actual mean of 52.5 degrees. Low Temperature (o F) 40 - 44 45 - 49 50 - 54 55 - 59 60 – 64 D Frequency 2 9. ... The mean of the frequency distribution is degrees. (Round to the nearest tenth as needed.) Which of the following best describes the relationship between the computed mean and the actual mean? O A. The computed mean is close to the actual mean because the difference between the means is more than 5% of the actual mean. B. The computed mean is not close to the actual mean because the difference between the means is less than 5% of the actual mean. C. The computed mean is not close to the actual mean because the difference between the means is more than 5% of the actual mean. D. The computed mean is close to the actual mean because the difference between the means is less than 5% of the actual mean.
What percent of students scored above 600? show your work.
Psychology students in the Izmir University of Economics completed the Dental Anxiety Scale questionnaire in 2020. In this study, scores on the scale range from 0 (no anxiety) to 20 (extreme anxiety). The mean score was 14 and the standard deviation was 3.5. Assume that the distribution of all scores on the Dental Arociety Scale is normal with u= 14 and o =3.5. Complete parts a through c A Click here to view a table of areas under the standardized normal curve A Table of Normal Curve Areas a. Suppose you score a 18 on the Dental Aniety Scale. Find the z-value for this score. z= (Round to the nearest hundredth as needed) b. Find the probability that someone scores between 13 and 18 on the Dental Anxiety Scale. P(13sxs 18) =O (Round to four decimal places as needed ) c. Find the probability that someone scores above 17 on the Dental Anxiety Scale Pix> 17) =D (Round to four decimal places as needed.) Enter your answer in each of the answer boxes 男明國 1 1 网解 8 1101 |8 与的 O

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