Answer:
A) 0.005
B) 0.001
C)0.0495
Step-by-step explanation:
Let A be the event that an aircraft is present and let B be the event the radar registers its presence.. Thus;
P(A) = Probability that an aircraft is present
P(A') = Probability that an aircraft is not present
P(B) = Probability that the radar generates an alarm
P(B') = Probability that the radar doesn't generate an alarm
Thus from what we are given, we have;
P(A) = 0.05
P(A') = 0.95
P(B) = 0.99
P(B') = 0.01
P(B|A') = 0.1
A) Probability of a false alarm will be;
P(A' ∩ B) = P(A) × P(B|A')
P(A' ∩ B) = 0.05 × 0.1 = 0.005
B) probability of missed detection is;
0.1 × (1 - 0.99) = 0.001
C) probability that an aircraft is present given that the radar registers a presence will be;
P(A ∩ B) = P(A) × P(B)
P(A ∩ B) = 0.05 × 0.99
P(A ∩ B) = 0.0495
Answer: The 95% confidence interval is approximately (55.57, 58.43)
======================================================
Explanation:
At 95% confidence, the z critical value is about z = 1.960 which you find using a table or a calculator.
The sample size is n = 17
The sample mean is xbar = 57
The population standard deviation is sigma = 3
The lower bound of the confidence interval is
L = xbar - z*sigma/sqrt(n)
L = 57 - 1.960*3/sqrt(17)
L = 55.5738905247863
L = 55.57
The upper bound is
U = xbar + z*sigma/sqrt(n)
U = 57 + 1.960*3/sqrt(17)
U = 58.4261094752137
U = 58.43
Therefore the confidence interval (L, U) turns into (55.57, 58.43) which is approximate.
<h3>
Answer: c = 29</h3>
Work Shown:
Use the pythagorean theorem
a^2+b^2 = c^2
21^2 + 20^2 = c^2
441 + 400 = c^2
841 = c^2
c^2 = 841
c = sqrt(841)
c = 29
