Answer:
The scale factor is 0.5
Step-by-step explanation:
Divide.
2/1=2
4/2=2
4/2=2
8/4=2
2/1=2
5/1.25=2
Now then, since this is a shrinking factor, divide 1 by the common outcome from the last section (2) to get 0.5
Answer:
a) P(X > 10) = 0.6473
b) P(X > 20) = 0.4190
c) P(X < 30) = 0.7288
d) x = 68.87
Step-by-step explanation:
Exponential distribution:
The exponential probability distribution, with mean m, is described by the following equation:
![f(x) = \mu e^{-\mu x}](https://tex.z-dn.net/?f=f%28x%29%20%3D%20%5Cmu%20e%5E%7B-%5Cmu%20x%7D)
In which
is the decay parameter.
The probability that x is lower or equal to a is given by:
![P(X \leq x) = \int\limits^a_0 {f(x)} \, dx](https://tex.z-dn.net/?f=P%28X%20%5Cleq%20x%29%20%3D%20%5Cint%5Climits%5Ea_0%20%7Bf%28x%29%7D%20%5C%2C%20dx)
Which has the following solution:
![P(X \leq x) = 1 - e^{-\mu x}](https://tex.z-dn.net/?f=P%28X%20%5Cleq%20x%29%20%3D%201%20-%20e%5E%7B-%5Cmu%20x%7D)
The probability of finding a value higher than x is:
![P(X > x) = 1 - P(X \leq x) = 1 - (1 - e^{-\mu x}) = e^{-\mu x}](https://tex.z-dn.net/?f=P%28X%20%3E%20x%29%20%3D%201%20-%20P%28X%20%5Cleq%20x%29%20%3D%201%20-%20%281%20-%20e%5E%7B-%5Cmu%20x%7D%29%20%3D%20e%5E%7B-%5Cmu%20x%7D)
Mean equal to 23.
This means that ![m = 23, \mu = \frac{1}{23} = 0.0435](https://tex.z-dn.net/?f=m%20%3D%2023%2C%20%5Cmu%20%3D%20%5Cfrac%7B1%7D%7B23%7D%20%3D%200.0435)
(a) P(X >10)
![P(X > 10) = e^{-0.0435*10} = 0.6473](https://tex.z-dn.net/?f=P%28X%20%3E%2010%29%20%3D%20e%5E%7B-0.0435%2A10%7D%20%3D%200.6473)
So
P(X > 10) = 0.6473
(b) P(X >20)
![P(X > 20) = e^{-0.0435*20} = 0.4190](https://tex.z-dn.net/?f=P%28X%20%3E%2020%29%20%3D%20e%5E%7B-0.0435%2A20%7D%20%3D%200.4190)
So
P(X > 20) = 0.4190
(c) P(X <30)
![P(X \leq 30) = 1 - e^{-0.0435*30} = 0.7288](https://tex.z-dn.net/?f=P%28X%20%5Cleq%2030%29%20%3D%201%20-%20e%5E%7B-0.0435%2A30%7D%20%3D%200.7288)
So
P(X < 30) = 0.7288
(d) Find the value of x such that P(X > x) = 0.05
So
![P(X > x) = e^{-\mu x}](https://tex.z-dn.net/?f=P%28X%20%3E%20x%29%20%3D%20e%5E%7B-%5Cmu%20x%7D)
![0.05 = e^{-0.0435x}](https://tex.z-dn.net/?f=0.05%20%3D%20e%5E%7B-0.0435x%7D)
![\ln{e^{-0.0435x}} = \ln{0.05}](https://tex.z-dn.net/?f=%5Cln%7Be%5E%7B-0.0435x%7D%7D%20%3D%20%5Cln%7B0.05%7D)
![-0.0435x = \ln{0.05}](https://tex.z-dn.net/?f=-0.0435x%20%3D%20%5Cln%7B0.05%7D)
![x = -\frac{\ln{0.05}}{0.0435}](https://tex.z-dn.net/?f=x%20%3D%20-%5Cfrac%7B%5Cln%7B0.05%7D%7D%7B0.0435%7D)
![x = 68.87](https://tex.z-dn.net/?f=x%20%3D%2068.87)
Answer:
201.06 ft
Step-by-step explanation:
Answer:
a. Yes. This provides convincing evidence that the true proportion of all attendees who ate the fish that got sick (80%) is more than the true proportion of all attendees who did not eat the fish that got sick.
b. The mistake here would have been the rejection of the Doctor's theory or hypothesis to the effect that more attendees who ate the fish got sick than those who did not eat the fish. This is a Type 1 error. A Type 1 error occurs when a null hypothesis is rejected when it is true. On the other hand, a Type II error occurs when the null hypothesis is accepted when it should be rejected. While a Type I error is equivalent to a false positive, a Type II error is equivalent to a false negative.
Step-by-step explanation:
Total number of attendees who ordered fish = 1,000
Sample size of the attendees who ate fish = 80
Number of attendees who ate the fish and got sick = 64 (80% or 64/80)
Sample size of attendees who did not eat fish = 60
Number of attendees who did not eat fish and got sick = 39 (65% or 39/60)
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