**Problem # 1**Assuming that the heights of college women are normally distributed with a mean of 66 inches and a standard deviation of 2.5 inches, answer the questions below.**1A.** Draw the **normal curve** to represent this data. Label the mean and the standard deviations along the top of the curve. Be sure to include a line along the bottom. Shade in the area that represents the probability that a college woman’s height will be between 58.5 inches and 68.5 inches.**1B.** What *percentage* of these women are taller than 68.5 inches? 1B. ___________**1C.** What *percentage* of these women are shorter than 61 inches? 1C. ___________**1D.** What is the *probability* that one of these women will be taller than 63.5 inches? 1D. ___________

Answered: Image /qna-images/answer/60a72a04-a441-4c36-ae20-fc67f72b3c12.jpg

Answered: Problem # 1 Assuming that the heights…

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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Refer to the figure at the right. At a =0.10, can you reject the claim that the proportion of 18 to 24-year-olds living in their parents' homes in 2000 was the same for men and women? Assume the survey included a random sample of 248 men and 244 women. Percentage of 18- to 24-year olds iving in parents' homes 7051.5% 41.2% 46.6% 44.3% 2000 2012 I Men I Women Identify the null and alternative hypotheses. Choose the correct answer below. O B. Ho: P1 P2 O D. Ho: P, > P2 Cc. Họ: P1 = P2 Ha: P, SP2 H3: P, *P2 Find the critical value(s). The critical value(s) is(are) +1.65. (Round to two decimal places as needed. Use a comma to separate answers as needed.) Find the standardized test statistic. z= 2.29 (Round to two decimal places as needed.) Can you reject the claim that the proportion of 18 to 24-year-olds living in their parents' homes was the same for men and women in 2000? Yes , at the 10% significance level, there is sufficient evidence to support the claim.
please explain how to find proportions and how I know which side the tail lands on.
SOLVE IN EXCEL PLEASE STEP BY STEP INSTRUCTIONS: Calculate the measures of central tendency (mean, median, and mode) and the measures of dispersion (range, mean, standard deviation, and variance) in the examples: 6. Find the range of spread and the population standard deviation of the following data: 3, 0, 4, 1, 2, 2, 1, 3, 3, 2, 2, 2, 3, 1, 1, 2, obtained during experimental measurements. Tip: We write down in a table the given data arranged in order ascending, the calculations and the results obtained: 6543210 0 1 2 3 4 0 Σ fl 1 4 =16 2 The arithmetic mean is: X = 0= 1 16 1 16 3 El=1*ifi 71 The population standard deviation is: The range of the dispersion is: T = Xmax-Xmin = 71 |2|\,(xr− x)?fi Σ= = x1/1 X 0 4 2 X₁ -2 2 (x₁-x)² 4 Σ= (x₁-x)²/1 Σ-
One graph in the figure represents a normal distribution with mean u= 15 and standard deviation o = 2. The other graph represents a normal distribution with mean u=11 and standard deviation o = 2. Determine which graph is which and explain how you know. A B 11 15 ... Choose the correct answer below. O A. Graph A has a mean of u = 15 and graph B has a mean of u = 11 because a larger mean shifts the graph to the right. O B. Graph A has a mean of u = 11 and graph B has a mean of u= 15 because a larger mean shifts the graph to the left. O C. Graph A has a mean of u = 11 and graph B has a mean of u = 15 because a larger mean shifts the graph to the right. O D. Graph A has a mean of u = 15 and graph B has a mean of u = 11 because a larger mean shifts the graph to the left.
One graph in the figure represents a normal distribution with mean = 15 and standard deviation o = 2. The other graph represents a normal distribution with mean u = 7 and standard deviation o = 2. Determine which graph is which and explain how you know .. Choose the correct answer below. OA. Graph A has a mean of μ = 15 and graph B has a mean of µ = 7 because a larger mean shifts the graph to the left. OB. Graph A has a mean of µ = 15 and graph B has a mean of μ = 7 because a larger mean shifts the graph to the right. OC. Graph A has a mean of u = 7 and graph B has a mean of μ = 15 because a larger mean shifts the graph to the right. G D. Graph A has a mean of μ = 7 and graph B has a mean of µ = 15 because a larger mean shifts the graph to the left. 7 15 B M
The scores and their percent of the final grade for a statistics student are given. What is the student's weighted mean score? The student's weighted mean score is. (Simplify your answer. Round to two decimal places as needed.) Homework Quiz Quiz Project Final Exam Score Percent of final grade 85 90 90 98 89 15 10 10 35 30
Think of your own example of a data set that would follow a normal distribution and explain why.
Psychology students at a university completed the Dental Anxiety Scale questionnaire. Scores on the scale range from 0 (no anxiety) to 20 (extreme anxiety). The mean score was 11 and the standard deviation was 3.5. Assume that the distribution of all scores on the Dental Anxiety Scale is normal with u = 11 and o = 3.5. Complete parts a through c. Click here to view a table of areas under the standardized normal curve. a. Suppose you score a 18 on the Dental Anxiety Scale. Find the z-value for this score. z= 2 (Round to the nearest hundredth as needed.) b. Find the probability that someone scores between 10 and 15 on the Dental Anxiety Scale. P(10sxs 15) = (Round to four decimal places as needed.) c. Find the probability that someone scores above 17 on the Dental Anxiety Scale. P(x> 17) =| (Round to four decimal places as needed.)
Please see attached.
d. Estimate the blood pressure of the student if his stress test score is 80, 60, and 85.
Please give me answer immediately.
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