Answer:
0.010
Step-by-step explanation:
We solve the above question using z score formula
z = (x-μ)/σ, where
x is the raw score = 63 inches
μ is the population mean = 70 inches
σ is the population standard deviation = 3 inches
For x shorter than 63 inches = x < 63
Z score = x - μ/σ
= 63 - 70/3
= -2.33333
Probability value from Z-Table:
P(x<63) = 0.0098153
Approximately to the nearest thousandth = 0.010
Therefore, the probability that a randomly selected student will be shorter than 63 inches tall, to the nearest thousandth is 0.010.
1/3 is equal to 33% because there are three different possibilties
Using the Poisson distribution, it is found that there is a 0.507 = 50.7% probability that the bird feeder will be visited by at most 5 birds in a 45 minute period during daylight hours.
<h3>What is the Poisson distribution?</h3>
In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by:
![P(X = x) = \frac{e^{-\mu}\mu^{x}}{(x)!}](https://tex.z-dn.net/?f=P%28X%20%3D%20x%29%20%3D%20%5Cfrac%7Be%5E%7B-%5Cmu%7D%5Cmu%5E%7Bx%7D%7D%7B%28x%29%21%7D)
The parameters are:
- x is the number of successes
- e = 2.71828 is the Euler number
is the mean in the given interval.
Considering the average of 15 birds every 2 hours during daylight hours, the mean for a 45-minute period is given by:
![\mu = 15 \times \frac{45}{120} = 5.625](https://tex.z-dn.net/?f=%5Cmu%20%3D%2015%20%5Ctimes%20%5Cfrac%7B45%7D%7B120%7D%20%3D%205.625)
The probability that the bird feeder will be visited by at most 5 birds in a 45 minute period during daylight hours is given by:
![P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)](https://tex.z-dn.net/?f=P%28X%20%5Cleq%205%29%20%3D%20P%28X%20%3D%200%29%20%2B%20P%28X%20%3D%201%29%20%2B%20P%28X%20%3D%202%29%20%2B%20P%28X%20%3D%203%29%20%2B%20P%28X%20%3D%204%29%20%2B%20P%28X%20%3D%205%29)
In which:
![P(X = x) = \frac{e^{-\mu}\mu^{x}}{(x)!}](https://tex.z-dn.net/?f=P%28X%20%3D%20x%29%20%3D%20%5Cfrac%7Be%5E%7B-%5Cmu%7D%5Cmu%5E%7Bx%7D%7D%7B%28x%29%21%7D)
![P(X = 0) = \frac{e^{-5.625}(5.625)^{0}}{(0)!} = 0.004](https://tex.z-dn.net/?f=P%28X%20%3D%200%29%20%3D%20%5Cfrac%7Be%5E%7B-5.625%7D%285.625%29%5E%7B0%7D%7D%7B%280%29%21%7D%20%3D%200.004)
![P(X = 1) = \frac{e^{-5.625}(5.625)^{1}}{(1)!} = 0.02](https://tex.z-dn.net/?f=P%28X%20%3D%201%29%20%3D%20%5Cfrac%7Be%5E%7B-5.625%7D%285.625%29%5E%7B1%7D%7D%7B%281%29%21%7D%20%3D%200.02)
![P(X = 2) = \frac{e^{-5.625}(5.625)^{2}}{(2)!} = 0.057](https://tex.z-dn.net/?f=P%28X%20%3D%202%29%20%3D%20%5Cfrac%7Be%5E%7B-5.625%7D%285.625%29%5E%7B2%7D%7D%7B%282%29%21%7D%20%3D%200.057)
![P(X = 3) = \frac{e^{-5.625}(5.625)^{3}}{(3)!} = 0.107](https://tex.z-dn.net/?f=P%28X%20%3D%203%29%20%3D%20%5Cfrac%7Be%5E%7B-5.625%7D%285.625%29%5E%7B3%7D%7D%7B%283%29%21%7D%20%3D%200.107)
![P(X = 4) = \frac{e^{-5.625}(5.625)^{4}}{(4)!} = 0.150](https://tex.z-dn.net/?f=P%28X%20%3D%204%29%20%3D%20%5Cfrac%7Be%5E%7B-5.625%7D%285.625%29%5E%7B4%7D%7D%7B%284%29%21%7D%20%3D%200.150)
![P(X = 5) = \frac{e^{-5.625}(5.625)^{5}}{(5)!} = 0.169](https://tex.z-dn.net/?f=P%28X%20%3D%205%29%20%3D%20%5Cfrac%7Be%5E%7B-5.625%7D%285.625%29%5E%7B5%7D%7D%7B%285%29%21%7D%20%3D%200.169)
Then:
![P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.004 + 0.02 + 0.057 + 0.107 + 0.15 + 0.169 = 0.507](https://tex.z-dn.net/?f=P%28X%20%5Cleq%205%29%20%3D%20P%28X%20%3D%200%29%20%2B%20P%28X%20%3D%201%29%20%2B%20P%28X%20%3D%202%29%20%2B%20P%28X%20%3D%203%29%20%2B%20P%28X%20%3D%204%29%20%2B%20P%28X%20%3D%205%29%20%3D%200.004%20%2B%200.02%20%2B%200.057%20%2B%200.107%20%2B%200.15%20%2B%200.169%20%3D%200.507)
0.507 = 50.7% probability that the bird feeder will be visited by at most 5 birds in a 45 minute period during daylight hours.
More can be learned about the Poisson distribution at brainly.com/question/13971530
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Answer:
Step-by-step explanation:
12x-34
Answer:
x-3 ,
Step-by-step explanation: