Calculate size of angle x. I worked out CA=14cm
1 answer:
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a) because it is equal to the area of the shaded region between X=4 and X=6, and the probability that X falls within some interval is given by the area under the PDF.
b) because the shaded region is a rectangle of height 1/5 (by virtue of X following a uniform distribution over the interval [2, 7], which has length 5).
Answer:
- The solution that optimizes the profit is producing 0 small lifts and 50 large lifts.
- Below are all the steps explained in detail.
- The graph is attached.
Explanation:
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<u>1. Name the variables:</u>
- x: number of smaller lifts
- y: number of larger lifts
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<u>2. Build a table to determine the number of hours each lift requires from each department:</u>
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Number of hours
small lift large lift total per department
Welding department 1x 3y x + 3y
Packaging department 2x 1y 2x + y
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<u>3. Constraints</u>
- 150 hours available in welding: x + 3y ≤ 150
- 120 hours available in packaging: 2x + y ≤ 120
- The variables cannot be negative: x ≥ 0, and y ≥ 0
Then you must:
- draw the lines and regions defined by each constraint
- determine the region of solution that satisfies all the constraints
- determine the vertices of the solution region
- test the profit function for each of the vertices. The vertex that gives the greatest profit is the solution (the number of each tupe that should be produced to maximize profits)
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<u>4. Graph</u>
See the graph attached.
Here is how you draw it.
- x + 3y ≤ 150
- draw the line x + 3y = 150 (a solid line because it is included in the solution set)
- shade the region below and to the left of the line
- 2x + y ≤ 120
- draw the line 2x + y ≤ 120 (a solid line because it is included in the solution set)
- shade the region below and to the left of the line
- x ≥ 0 and y ≥ 0: means that only the first quadrant is considered
- the solution region is the intersection of the regions described above.
- take the points that are vertices inside the solutoin region.
<u>5. Test the profit function for each vertex</u>
The profit function is P(x,y) = 25x + 90y
The vertices shown in the graph are:
- (0,0)
- (0,50)
- (42,36)
- (60,0)
The profits with the vertices are:
- P(0,0) = 0
- P(0,50) = 25(0) + 90(50) = 4,500
- P(42,36) = 25(42) + 90(36) = 4,290
- P(60,0) = 25(60) + 90(0) = 1,500
Thus, the solution that optimizes the profit is producing 0 smaller lifts and 90 larger lifts.
Answer:
D. 0.9953 (Probability of a Type II error), 0.0047 (Power of the test)
Step-by-step explanation:
Let's first remember that a Type II error is to NOT reject H0 when it is false, and the probability of that occurring is known as β. On the other hand, power refers to the probability of rejecting H0 when it is false, so it can be calculated as 1 - β.
To resolve this we are going to use the Z-statistic:
Z = (X¯ - μ0) / (σ/√n)
where μ0 = 1.2
σ = 5
n = 42
As we can see in part A of the attached image, we have the normal distribution curve representation for this test, and because this is a two-tailed test, we split the significance level of α=0.01 evenly into the two tails, 0.005 in each tail, and if we look for the Z critical value for those values in a standard distribution Z table we will find that that value is 2.576.
Now we need to stablish the equation that will telll us for what values of X¯ will we reject H0.
Reject if:
Z ≤ -2.576 Z ≥ 2.576
We know the equation for the Z-statistic, so we can substitute like follows and resolve.
Reject if:
(X¯ - 1.2) / (5/√42) ≤ -2.576 (X¯ - 1.2) / (5/√42) ≥ 2.576
X¯ ≤ -0.79 X¯ ≥ 3.19
We have the information that the true population mean is 1.25, so we now for a fact that H0 is false, so with this we can calculate the probability of a Type II error: P(Do not reject H0 | μ=1.25)
As we can see in part B of the attached image, we can stablish that the type II error will represent the probability of the sample mean (X¯) falling between -0.79 and 3.19 when μ=1.25, and that represents the shaded area. So now we now that we are looking for P(-0.79 < X¯ < 3.19 | μ=1.25).
Because we know the equation of Z, we are going to standardize this as follows:
P ( (-0.79 - 1.25) / (5/√42) < Z < (3.19 - 1.25) / (5/√42) )
This equals to:
P(-2.64 < Z < 2.51)
If we go and look for the area under the curve for Z positive scores in a normal standart table (part C of attached image), we will find that that area is 0.9940, which represents the probability of a Type II error.
Therefore, the power of the test will be 1-0.9940 = 0.006
If we look at the options of answers we have, there is no option that looks like this results, which means there was a probable redaction error, so we are going to stay with the closest option to these values which is option D.
