A rental car company charges $68.64 per day to rent a car and $0.12 for every mile driven. Jaxon wants to rent a car, knowing th
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Answer:
0.0588 = 5.88% probability that a middle-aged man with diabetes is very active
Step-by-step explanation:
Conditional Probability
We use the conditional probability formula to solve this question. It is
In which
P(B|A) is the probability of event B happening, given that A happened.
is the probability of both A and B happening.
P(A) is the probability of A happening.
In this question:
Event A: Has diabetes.
Event B: Is very active.
Probability of having diabetes:
To find this probability, we take in consideration that:
It also found that men who were very active (burning about 3,500 calories daily) were a fourth as likely to develop diabetes compared with men who were sedentary. Assume that one-fifth of all middle-aged men are very active, and the rest are classified as sedentary.
So the probability of developing diabetes is:
x of 4/5 = x of 0.8(not active)
x/4 = 0.25x of 1/5 = 0.2(very active). So
Probability of developing diabetes while being very active:
0.25x of 0.2. So
What is the probability that a middle-aged man with diabetes is very active?
0.0588 = 5.88% probability that a middle-aged man with diabetes is very active
Answer:
The correct answer is
(0.0128, 0.0532)
Step-by-step explanation:
In a sample with a number n of people surveyed with a probability of a success of , and a confidence interval
, we have the following confidence interval of proportions.
In which
Z is the zscore that has a pvalue of
For this problem, we have that:
In a random sample of 300 circuits, 10 are defective. This means that and
Calculate a 95% two-sided confidence interval on the fraction of defective circuits produced by this particular tool.
So = 0.05, z is the value of Z that has a pvalue of
, so
.
The lower limit of this interval is:
The upper limit of this interval is:
The correct answer is
(0.0128, 0.0532)