(d) In the sample obtained in part (a), what is the probability that the proportion who are satisfied with the way things aregoing in their life exceeds 0.91?The probability that the proportion who are satisfied with the way things are going in their life exceeds 0.91 is(Round to four decimal places as needed.)(e) Using the distribution from part (c), would it be unusual for a survey of 100 people to reveal that 81 or fewer people inthe sample are satisfied with their lives?The probability that 81 or fewer people in the sample are satisfied isless than%.(Round to four decimal places as needed.)whichunusual because this probability

Answered: Image /qna-images/answer/20b85773-b7f8-44c5-9705-54676dfc4f57.jpg

Answered: (d) In the sample obtained in part (a),…

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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About 79% of all female heart transplant patients will survive for at least 3 years. Eighty female heart transplant patients are randomly selected. What is the probability that the sample proportion surviving for at least 3 years will be less than 70%? Assume the sampling distribution of sample proportions is a normal distribution. pq The mean of the sample proportion is equal to the population proportion and the standard deviation is equal to n The probability that the sample proportion surviving for at least 3 years will be less than 70% is| (Round to four decimal places as needed.)
2
Based on historical data, your manager believes that 27% of the company's orders come from first- time customers. A random sample of 74 orders will be used to estimate the proportion of first-time- customers. Show your answer to 4 decimal places if necessary. a. What is the distribution of p? p ~ N( b. What is the probability that the sample proportion is less than 0.25? c. What is the probability that the sample proportion is between 0.22 and 0.3? Add Work MacBook DII F10 ה כ F9 F8 F6 F7 F5 F4 F3 F2 F1 & # 2$ % ! @ 7 8. 3 4 5 1 U T
About 71% of all female heart transplant patients will survive for at least 3 years. Eighty female heart transplant patients are randomly selected. What is the probability that the sample proportion surviving for at least 3 years will be pq less than 63%? Assume the sampling distribution of sample proportions is a normal distribution. The mean of the sample proportion is equal to the population proportion and the standard deviation is equal to The probability that the sample proportion surviving for at least 3 years will be less than 63% is (Round to four decimal places as needed.) More Clear All Final Check Help Me Solve This View an Example Get More Help - 4:51 PM 73°F Rain coming 4* 10/4/2021 O Type here to search 10/05/17 EGO PrtSc Insert Delete 口回 F10 F11 F12 F7 F8 F9 F6 Esc F2 F4 F5 F1 F3 & Backspace Num Lock 5 6 2 R T. Y U Home G K L Enter F H. J + |I LEGY P * 00 44 %#3 山

In a normal distribution of scores, four participants obtained the following deviation scores: +1, −2, +5, and −10. (a) Which score reflects the highest raw score? (b) Which score reflects the lowest raw score? (c) Rank-order the deviation scores in terms of their frequency, starting with the score with the lowest frequency.
Answer D
Do you think new package labeling or doing nothing is best? You will want to calculate the percent of bars produced that are within 5 grams of 1000 grams and the percent of bars produced that are within 5 grams of 990 grams. In other words, given a uniform distribution between 980 and 1000, what is the probability of being between 995 and 1005 and what is the probability of being between 985 and 995?
Answer A-C
Is the difference between the mean annual salaries of statisticians in Region 1 and Region 2 more than $5000? To decide, you select a random sample of statisticians from each region. The results of each survey are shown to the right. At a = 0.10, what should you conclude? Region 1 X1 = $67,700 Region 2 X2 = $62,000 01 = $8825 02 = $9025 n1 = 48 n2 = 43 Click here to view page 1 of the standard normal distribution table. Click here to view page 2 of the standard normal distribution table. ..... Choose the correct null and alternative hypotheses below. B. Ho: H1- H2 O C. Ho: H1 -H2 > 5000 Ha: H1 - H2 55000 O A. Ho: H1 - H2 > 5000 + 5000 Ha: H1 - H2 2 5000 Ha: H1 - H2 = 5000 O E. Ho: H1 - H2 2 5000 O D. Ho: H1 - H2 = 5000 Ha: H1 - H2 # 5000 F. Ho: µ1 - H2 5 5000 Ha: H1 - H2 > 5000 Ha: H1 - H2 В. Z
r a standardized psychology examination intended for psychology majors, the historical data show that scores have a mean of 510 and a standard deviation of 170 . The grading process of this year's exam has just begun. The average score of the 40 exams graded so far is 526 . What is the probability that a sample of 40 exams will have a mean score of 526 or more if the exam scores follow the same distribution as in the past? Carry your intermediate computations to at least four decimal places. Round your answer to at least three decimal places.
Is the difference between the mean annual salaries of statisticians in Region 1 and Region 2 more than $4000? To decide, you select a random sample of statisticians from each region. The results of each survey are shown to the right. At a = 0.10, what should you conclude? Region 1 X1 = $67,700 01 = $8875 Region 2 X2 = $61,000 02 = $9250 n1 = 48 n2 = 44 Click here to view page 1 of the standard normal distribution table. Click here to view page 2 of the standard normal distribution table. Choose the correct answer below. O A. Reject Ho. At the 10% significance level, there is sufficient evidence to support the claim that the difference between the mean annual salaries is more than $4000. O B. Fail to reject Ho. At the 10% significance level, there is sufficient evidence to support the claim that the difference between the mean annual salaries is more than $4000. OC. Reject Ho. At the 10% significance level, there is insufficient evidence to support the claim that the difference between…
6. A study was conducted to determine the number of emergency calls per day received by the Conestoga Valley Fire Company. The data are shown here: # of phone calls Frequency 5 7 8 28 32 21 10 (a) Construct a probability distribution for the number of phone calls. (b) Find the mean number of phone calls, rounded to the nearest tenth.

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