. 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
. In one town, 67% of adults have health insurance. What is the probability that 5...
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, 61% of adults have health insurance. What is the probability that 6 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
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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?
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 step by step and round to 4 decimal places.
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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...
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)
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....
equen 36% of working mothers do not have enough money to cover their health insurance deductibles. You randomly select six working mothers and ask them whether they have enough money to cover their health insurance deductibles. The random variable represents the number of working mothers who do not have enough money to their health insurance deductibles. Complete parts (a) through (c) below (a) Construct a binomial distribution using n 6 and p 0.36 P(x 0 2 For H 4 5...
Use the given degree of confidence and sample data to construct a confidence interval for the population proportion p. Round to three decimal palces Of 98 adults selected random from one town, 68 have health insurance Find a 90% confidence interval adults in he own who have heal e proportion on r e a ns rance 0585 < p < 0 802 B. 0.617<p<0.770 A. C. 0603 p<0.785 D. 0.574p<0.814
Use the given degree of confidence and sample data to...