My dear student I am allow to solve the one question at a time. So I solved the first question in the picture.
Provide an appropriate response. Use the Standard NormalTable to find the probably The distribution of cholesterol...
Find the mean, variance, and standard deviation of the binomial distribution with the given values of n and p. n = 121, p=0.32 The mean, H, is (Round to the nearest tenth as needed.) The variance, 02, is (Round to the nearest tenth as needed.) The standard deviation, o, is (Round to the nearest tenth as needed.) 47% of U.S. adults have very confidence in newspapers. You randomly select 10 US adults. Find the probably at the number of US...
Provide an appropriate response. Use the Standard Normal Table to find the probability. 11) The distribution of cholesterol levels in teenage boys is approximately normal with 170 ando - 30 (Source: U.S. National Center for Health Statistics). Levels above 200 warrant attention. Find the probability that a teenage boy has a cholesterol level greater than 225. A) 0.0718 B) 0.0012 C) 0.0606 D) 0.0336
46% of U.S. adults have very little confidence in newspapers. You randomly select 10 U.S. adults. Find the probability that the number of U.S. adults who have very little confidence in newspapers is (a) exactly five, (b) at least six, and (c) less than four. 46% of U.S. adults have very little confidence in newspapers. You randomly select 10 U.S. adults. Find the probability that the number of U.S. adults who have very little confidence in newspapers is (a)...
56% of U.S. adults have very little confidence in newspapers. You randomly select 10 U.S. adults. Find the probability that the number of U.S. adults who have very little confidence in newspapers is (a) exactly five, (b) at least six, (c) less than four. (Round to three decimal places as needed.)
69% of U.S. adults have very little confidence in newspapers. You randomly select 10 U.S. adults. Find the probability that the number of U.S. adults who have very little confidence in newspapers is (a) exactly five, (b) at least six, and (c) less than four. (a) P(5) = (Round to three decimal places as needed.) (b) P(x26) = (Round to three decimal places as needed.) (c) P(x<4) =(Round to three decimal places as needed.)
15 Question Help 47% of U.S. adults have very little confidence in newspapers. You randomly select 10 U.S. adults. Find the probability that the number of U.S. adults who have very little confidence in newspapers is (a) exactly five, (b) at least six, and (c) less than four (a) P(5)=L(Round to three decimal places as needed.) (b) P(x26) = (Round to three decimal places as needed.) (c) P(x<4)= (Round to three decimal places as needed.) Enter your answer in each...
Find the probability that the number of u.s adults who have very little confidence in newspapers is (a)exactly 5 (b) AT LEAST 6 (c) LESS THAN FOUR core: 0 of 1 pt 7of11(6complete) ▼ HW Score: 54.55%, 6 of 11 2.19 Question Help n newspapers. You randomly select 10 U.S. aduts. Find the probability that the number of U.S. adults whe have very litle conidence in newepapers is (a) exactly five, (b) at least six, and (e)le han four PS(Round...
In a best of the effectiveness of garlic for lowering cholesterol 43 wucts were treated with garlic in a processed abietom. Cholesterol levels were measured before and the rest. The changes before her in the levels of LOL cholesterol in my have a mean of 3.6 and a standard deviation of 152 Construct a 10% confidence Wierval estate of the mean not change in LDL cholesterol to the garlic treatment. What does the confidence interval suggest about the effectiveness of...
Use the given degree of confidence and sample data to construct a confidence interval for the population proportion p. Round to three decimal palces. or 97 adults selected randomly from one town, B4 have health insurance. Find a 90% confidence interval for the true proportion of all adults in the town who have health insurance O A. 0.581 <p <0.739 OB. 0.566 <p <0.754 OC. 0.536 <p<0.784 OD. 0.548<p<0.772
Use the given degree of confidence and sample data to construct a confidence interval for the population proportion p. Round to three decimal places When 310 college students are randomly selected and surveyed, it is found that 112 own a car. Find a 99% confidence interval for the true proportion of all college students who own a car. O A. 0.291 <p<0.432 OB. 0.298 <p<0.425 OC. 0.316<p<0.406 OD. 0.308<p <0.415