If the probability of being hospitalized during a year is 0.15 find the probability that no one in the family of 4 will be hospi
1 answer:
The probability that no one in the family of 4 will be hospitalized during a year is 0.52
Step-by-step explanation:
The probability for a person to be hospitalized during one year is
As a consequence, the probability for a person of NOT being hospitalized during one year is
If we have 4 people in the family, the probability of each of them of NOT being hospitalized during the year is independent from the probability of the others - therefore, these are independent events. This means that the total probability of the 4 people of not being hospitalized during a year is:
So, the probability that no one in the family of 4 will be hospitalized during a year is 0.52.
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Answer:
A sample size of 345 is needed so that the confidence interval will have a margin of error of 0.07
Step-by-step explanation:
In a sample with a number n of people surveyed with a probability of a success of , and a confidence level of
, we have the following confidence interval of proportions.
In which
z is the zscore that has a pvalue of .
The margin of error of the interval is given by:
In this problem, we have that:
99.5% confidence level
So , z is the value of Z that has a pvalue of
, so
.
Using this estimate, what sample size is needed so that the confidence interval will have a margin of error of 0.07?
This is n when M = 0.07. So
A sample size of 345 is needed so that the confidence interval will have a margin of error of 0.07