The probability according to the given scenario is:
(a) <em>0.0395</em>
(b) <em>0.95</em>
(c) <em>0.076</em>
The given values are:
(a)
The probability that the b/w has tuberculosis as well as having (+) test will be:
→ 
By substituting the values, we get
→ 
→ 
(b)
The probability that a person does not have tuberculosis will be:
→ 
→ 
→ 
(c)
The person does not have tuberculosis as well as have a (+) test, the probability will be:
→ 
→ 
→ 
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Answer:
A sample of 179 is needed.
Step-by-step explanation:
We have that to find our 
level, that is the subtraction of 1 by the confidence interval divided by 2. So:
Now, we have to find z in the Ztable as such z has a pvalue of 
.
That is z with a pvalue of 
, so Z = 1.44.
Now, find the margin of error M as such
In which 
is the standard deviation of the population and n is the size of the sample.
A previous study found that for an average family the variance is 1.69 gallon?
This means that 
If they are using a 85% level of confidence, how large of a sample is required to estimate the mean usage of water?
A sample of n is needed, and n is found for M = 0.14. So
Rounding up
A sample of 179 is needed.
First one m = .0666666
second one i don't know
third one is .1
5(t)
It will be $5 times however many weeks
