Find each uniform continuous probability and sketch a graph showing it as a shaded area.. If you please show me how you got the answer I would really appreciate.
a. P(X < 10) for U (0, 50)
b. P(X > 500) for U (0, 1,000)
c. P(25< X < 45) for U(15, 65)
a. To find the probability P(X < 10) for a uniform continuous distribution on the interval (0, 50), we need to calculate the area under the probability density function (PDF) curve for values less than 10.
The PDF for a uniform continuous distribution on the interval (0, 50) is a horizontal line with height 1/50, since the interval has a length of 50 and the total area under the PDF curve must be equal to 1.
To find the shaded area representing P(X < 10), we need to calculate the area of a rectangle with width 10 and height 1/50. The area is given by the formula: Area = width * height. In this case, the area is (10 * 1/50) = 1/5.
So, P(X < 10) = 1/5.
Here's a sketch of the graph:
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0 10 50
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1/5
b. To find the probability P(X > 500) for a uniform continuous distribution on the interval (0, 1000), we need to calculate the area under the PDF curve for values greater than 500.
The PDF for a uniform continuous distribution on the interval (0, 1000) is a horizontal line with height 1/1000, since the interval has a length of 1000 and the total area under the PDF curve must be equal to 1.
To find the shaded area representing P(X > 500), we need to calculate the area of a rectangle with width 500 and height 1/1000. The area is given by the formula: Area = width * height. In this case, the area is (500 * 1/1000) = 1/2.
So, P(X > 500) = 1/2.
Here's a sketch of the graph:
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0 500 1000
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1/2
c. To find the probability P(25 < X < 45) for a uniform continuous distribution on the interval (15, 65), we need to calculate the area under the PDF curve between 25 and 45.
The PDF for a uniform continuous distribution on the interval (15, 65) is a horizontal line with height 1/50, since the interval has a length of 50 and the total area under the PDF curve must be equal to 1.
To find the shaded area representing P(25 < X < 45), we need to calculate the area of a rectangle with width 20 and height 1/50. The area is given by the formula: Area = width * height. In this case, the area is (20 * 1/50) = 2/5.
So, P(25 < X < 45) = 2/5.
Here's a sketch of the graph:
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15 25 45 65
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2/5
To find each uniform continuous probability and sketch the graph showing it as a shaded area, we will use the formula for the probability density function (PDF) of a uniform distribution:
f(x) = 1 / (b - a)
where "a" is the lower limit of the distribution and "b" is the upper limit.
a. P(X < 10) for U (0, 50):
First, we need to find the probability density function (PDF) for this uniform distribution:
f(x) = 1 / (50 - 0) = 1 / 50
To find P(X < 10), we integrate the PDF from 0 to 10:
P(X < 10) = ∫[0 to 10] f(x) dx = ∫[0 to 10] 1 / 50 dx
Integrating, we get:
P(X < 10) = [1 / 50 * x] [0 to 10] = (1 / 50 * 10) - (1 / 50 * 0) = 1 / 5 - 0 = 1/5 = 0.2
To sketch the graph, you can draw the x-axis from 0 to 50 and shade the area under the curve up to x = 10.
b. P(X > 500) for U (0, 1,000):
Similarly, for this uniform distribution, the PDF is:
f(x) = 1 / (1,000 - 0) = 1 / 1,000
To find P(X > 500), we integrate the PDF from 500 to 1,000:
P(X > 500) = ∫[500 to 1,000] f(x) dx = ∫[500 to 1,000] 1 / 1,000 dx
Integrating, we get:
P(X > 500) = [1 / 1,000 * x] [500 to 1,000] = (1 / 1,000 * 1,000) - (1 / 1,000 * 500) = 1 - 0.5 = 0.5
To sketch the graph, draw the x-axis from 0 to 1,000 and shade the area under the curve starting from x = 500 to the right.
c. P(25 < X < 45) for U(15, 65):
Using the same formula, the PDF for this distribution is:
f(x) = 1 / (65 - 15) = 1 / 50
To find P(25 < X < 45), we integrate the PDF from 25 to 45:
P(25 < X < 45) = ∫[25 to 45] f(x) dx = ∫[25 to 45] 1 / 50 dx
Integrating, we get:
P(25 < X < 45) = [1 / 50 * x] [25 to 45] = (1 / 50 * 45) - (1 / 50 * 25) = 9/10 - 1/2 = 0.9 - 0.5 = 0.4
To sketch the graph, draw the x-axis from 15 to 65 and shade the area under the curve between x = 25 and x = 45.
Sure! To find the probabilities for each uniform continuous probability distribution, we can use the properties of the uniform distribution.
For a uniform continuous distribution U(a, b), where 'a' is the lower limit and 'b' is the upper limit, the probability density function (PDF) is given by:
f(x) = 1 / (b - a) for a ≤ x ≤ b
0 otherwise
Now, let's calculate the probabilities and sketch the graphs for each question:
a. P(X < 10) for U (0, 50):
To find this probability, we need to calculate the area under the PDF curve from 0 to 10.
The PDF in this case is: f(x) = 1 / (50 - 0) = 1 / 50
To find the probability, we integrate the PDF over the interval from 0 to 10:
P(X < 10) = ∫[0,10] (1 / 50) dx
= (1 / 50) * [x] from 0 to 10
= (1 / 50) * (10 - 0)
= 10 / 50
= 1 / 5
= 0.2
So, P(X < 10) for U (0, 50) is 0.2.
b. P(X > 500) for U (0, 1,000):
Similar to the previous case, we need to find the area under the PDF curve from 500 to 1,000.
The PDF in this case is: f(x) = 1 / (1,000 - 0) = 1 / 1,000
To find the probability, we integrate the PDF over the interval from 500 to 1,000:
P(X > 500) = ∫[500,1,000] (1 / 1,000) dx
= (1 / 1,000) * [x] from 500 to 1,000
= (1 / 1,000) * (1,000 - 500)
= 500 / 1,000
= 0.5
So, P(X > 500) for U (0, 1,000) is 0.5.
c. P(25 < X < 45) for U(15, 65):
To find this probability, we need to calculate the area under the PDF curve from 25 to 45.
The PDF in this case is: f(x) = 1 / (65 - 15) = 1 / 50
To find the probability, we integrate the PDF over the interval from 25 to 45:
P(25 < X < 45) = ∫[25,45] (1 / 50) dx
= (1 / 50) * [x] from 25 to 45
= (1 / 50) * (45 - 25)
= 20 / 50
= 0.4
So, P(25 < X < 45) for U(15, 65) is 0.4.
I hope this explanation and the sketches help you understand how to find the probabilities for uniform continuous probability distributions!