Answer:
In a sample of 1000 test the expected number of false negatives is 8.6.
Step-by-step explanation:
The table provided is:
Positive Negative TOTAL
Hepatitis C 335 10 345
No Hepatitis C 2 1153 1155
TOTAL 337 1163 1500
A false negative test result implies that, the person's test result for Hepatitis C was negative but actually he/she had Hepatitis C.
Compute the probability of false negative as follows:
![P(Negative|No\ Hepapatits\ C)=\frac{n(Negative\cap No\ Hepapatits\ C)}{n(No\ Hepapatits\ C)} \\=\frac{10}{1163}\\ =0.0086](https://tex.z-dn.net/?f=P%28Negative%7CNo%5C%20Hepapatits%5C%20C%29%3D%5Cfrac%7Bn%28Negative%5Ccap%20No%5C%20Hepapatits%5C%20C%29%7D%7Bn%28No%5C%20Hepapatits%5C%20C%29%7D%20%5C%5C%3D%5Cfrac%7B10%7D%7B1163%7D%5C%5C%20%3D0.0086)
Compute the expected number of false negatives in a sample is <em>n</em> = 1000 tests as follows:
E (False negative) = <em>n</em> × P (Negative|No Hepatitis C)
![=1000\times0.0086\\=8.6](https://tex.z-dn.net/?f=%3D1000%5Ctimes0.0086%5C%5C%3D8.6)
Thus, in a sample of 1000 test the expected number of false negatives is 8.6.
Answer:
k= 12 x= 5
Step-by-step explanation:
Answer:
6745.09
Step-by-step explanation:
If we assume the nominal annual interest rate is 5%, then the future value after 6 years is ...
FV = P(1 +r/12)^(12·t)
for P = 5000, r = .05, t = 6.
Doing the arithmetic, we get ...
FV = 5000(1 +.05/12)^(12·6) ≈ 6745.09
After 6 years, the bank account will be worth 6745.09.
_____
We made a comment about the interest rate, because the annual <em>yield</em> is about 5.116%. If the <em>annual yield</em> is actually 5%, then the account value is lower: $6700.48. (Monthly compounding is irrelevant in that case, because it is already figured into the yield.)
Usually, the wording would be that the account <em>earns</em> 5% interest compounded monthly.
Answer
9/20
Step-by-step explanation:
This is a probabilty, problem
firstly, let us bring out the sample space and size
the sample space is
9 apples,
5 bananas,
1 orange, and
5 peaches
The sample size is (9+5+1+5)= 20
The probability of picking a banana is
Pr(banana) =5/20
=1/4
The probability of picking an apple
Pr(apple) = 9/20
The probability of picking up a banana and an apple with replacement is
=1/4+9/20
=9/80
just multiple the numbers.
for example 7×7 is - 49y
and 3×16 is - 48y