Find the indicated probability In one town, 77% of adults have health insurance What is the probability that 10 adults selected at random from the town all have health insurance? Round to the nearest thousandth if necessary
. In one town, 67% of adults have health insurance. What is the probability that 5 adults selected at random from the town all have health insurance? Round to the nearest thousandth if necessary. A) 0.135 B) 0.075 C) 0.670 D) 3.350
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
Please show all work in detail, thank you! 9. In one town, 66% of adults have health insurance. What is the probability that 4 adults selected at random from the town ALL have health insurance?
1-Prop Conf. Int. 07 or 94 adults selected randomly from one town, 67 have health insurance. Find a 90% confidence interval for the true proportion of all adults in the town who have health insurance. Use Interval Notation with decimal rounded to the thousandths, Answer
5. In Salinas, 68% of adults have health insurance. If 400 adults were randomly selected. a) Find the probability that at least 230 of them have health insurance. (Use binomial approximation) b) Find the probability that exactly 230 of them have health insurance. (Use binomial approximation)
Please show all work step by step and round to 4 decimal places. thank you. 8. Of the 75 people who said they have sleepwalked, 58 were female. Of the 67 people who said they have not sleepwalked, 34 were male. If one person is selected at random from the group, find the probability that the randomly selected person was a female or has not sleepwalked. 9. In one town, 66% of adults have health insurance. What is the probability...
In a sample of 88 adults selected randomly from one town, it is found that 6 of them have been exposed to a particular strain of the flu. At the 0.01 significance level, test the claim that the proportion of all adults in the town that have been exposed to this strain of the flue differs from the nationwide percentage of 8%. H0: p = 0.08 HA: p ≠ 0.08. α = 0.01 Test statistic: z = -0.41. P-Value =...
of 146 adults selected randomly from one town, 25 of them smoke. Construct a 99% confidence interval for the true percentage of all adults in the town that smoke. Use the given degree of confidence and sample data to construct a confidence interval for the population proportion p. Points: 5 5) Of 146 adults selected randomly from one town, 25 of them smoke. Construct a 99% confidence interval for the true percentage of all adults in the town that smoke....
Use the given degree of confidence and sample data to construct a confidence interval for the population proportion p. Round to three decimal places. Of 89 adults selected randomly from one town, 66 have health insurance. Find a 90% confidence interval for the true proportion of all adults in the town who have health insurance.