TOPIC: Use of Binomial distribution.
So,the correct answer is- option-b> (1)0.247 (2) 0.744 (ANSWER)
Sixty-eight percent of US adults have ittle confidence in their cars. You randomly select eleven US...
Fifty-nine percent of US adults have little confidence in their cars. You randomly select eight US adults. Find the probability that the number of US adults who have little confidence in their cars is (1) exactly three and then find the probability that it is (2) more than 6. show the work please
Thirty-five percent of US adults have little confidence in their cars. You randomly select ten US adults. Find the probability that the number of US adults who have little confidence in their cars is (1) exactly six and then find the probability that it is (2) more than 7. (1) 0.069 (2) 0.005 (1) 0.069 (2) 0.974 (1) 0.021 (2) 0.005 (1) 0.021 (2) 0.026
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)...
42 % 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.
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...
Of 92 adults selected randomly from one town, 61 have health insurance. Find a 90% confidence interval for the true proportion of all adults in the town who have health insurance. 0.566 < p < 0.760 0.548 < p < 0.778 0.536 < p < 0.790 0.582 < p < 0.744
1/ Consider the following table. Defects in batch Probability 2 0.18 3 0.29 4 0.18 5 0.14 6 0.11 7 0.10 Find the standard deviation of this variable. 1.52 4.01 1.58 2.49 2/ The standard deviation of samples from supplier A is 0.0841, while the standard deviation of samples from supplier B is 0.0926. Which supplier would you be likely to choose based on these data and why? Supplier B, as their standard deviation is higher and, thus, easier to...