A contractor has found that her cost for a certain construction job is subject to random variation. She believes its actual valu
e follows a continuous uniform distribution between $9000 and $11000 with a mean of $10,000. The cost to prepare her bid is $500. She is competing in a sealed bid competition with four other contractors. She believes that the bids submitted by the other contractors will vary triangularly with a minimum equal to her minimum cost, a most likely value equal to her 1.3 times her mean cost and a maximum equal to 2.5 times her mean cost and that their bids are independent of each other. She will win the competition if her bid is smaller than the bids of all four competitors. If she wins, her profit will be the difference between her bid and her cost minus the proposal preparation cost. If she loses her proposal preparation cost is lost and her profit is -$500. Simulate 1000 bids in which her bid amount is $14,000 and determine the distribution of her profit. Include the histogram in your response. Also determine the probability that she wins the competition and the probability that she loses money. Compare these results for bid amounts of $13000 and $15000
1 answer:
Answer:
Attached is the profit distribution plotted on the chart and also the detailed solution using excel
Step-by-step explanation:
To solve this problem we have to
- create a column for number of counts ( 1,2,3........1000) bids
- create a column for the cost to be incurred which is mostly dependent on the random number generated. the formula for that using excel is; 9000+ rand()*(11000-9000) for uniform distribution between the numbers
- Four(4) more columns are generated for bids of competitors by using the formula: 10000+rand()*(3*10000-10000) this because the bids that will be submitted by others bidders will vary uniformly between her mean cost and 3 times her mean cost
- Condition is checked to see if the lowest bid is. =IF(MIN(the 4 bids)>14000,1,0)
- Next the same process is carried out for 13000 and 15000
- The probability of winning is calculated in excel using this formula =Countif(value of step 4 for all the rows,1)/1000
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To convert from rectangular to polar we will use these 2 formulas:
and
.
The r value found serves as the first coordinate in our polar coordinate, and the angle serves as the second coordinate of the pair. We are told to find 2. Since the r value will always be the same (it's the length of the hypotenuse created in the right triangle we form when determining our angle theta), the angle is what is going to be different in our coordinate pairs. We use the x and y coordinates from the given rectangular coordinate to solve for the r in both our coordinate pairs.
which gives us an r value of
. That's r for both coordinate pairs. Now we move to the angle. Setting up according to our formula we have
.
This asks the question "what angle(s) has/have a tangent of -1?". That's what we have to find out! Since the tangent ratio is y/x AND since it is negative, it is going to lie in a quadrant where x is negative and y is positive, AND where x is positive and y is negative. Those quadrants are 2 and 4. In QII, x is negative so the tangent ratio is negative here; in QIV, y is negative so the tangent ratio is negative here as well. Now, if we type inverse tangent of -1 into our calculators in degree mode, we get that the angle that has a tangent of -1 is -45. Measured from the positive x axis, -45 does in fact go into the fourth quadrant. However, since the inverse tangent of -1 is -45, we also have a 45 degree angle in the second quadrant. Those are reference angles, mind you. A 45 degree angle in QII has a coterminal angle of 135 degree; a 45 degree angle in QIV has a coterminal angle of 315. If you don't understand that, go back to your lesson on reference angles and coterminal angles to see what those are. So our polar coordinates for that rectangular coordinate are
and
The squares are shown in the attached picture.
As you can see, JC is half the diagonal of ABCD and JH is half the diagonal of EFGH.
In order to find the area of the shaded figure, we need to subtract the area of the white square (EFGH) from the area of the big square (ABCD).
The area of a square know the diagonal is given by the formula:
A = d² ÷ 2
A(ABCD) = (2×JC)² ÷ 2
= (2×9)² ÷ 2
= 162 cm²
A(EFGH) = (2×JH)² ÷ 2
= (2×4)² ÷ 2
= 32 cm²
Therefore:
A = A(ABCD) - <span>A(EFGH)
= 162 - 32
= 130 cm</span>²
The area of the shaded region is 130 cm².
As you can see, JC is half the diagonal of ABCD and JH is half the diagonal of EFGH.
In order to find the area of the shaded figure, we need to subtract the area of the white square (EFGH) from the area of the big square (ABCD).
The area of a square know the diagonal is given by the formula:
A = d² ÷ 2
A(ABCD) = (2×JC)² ÷ 2
= (2×9)² ÷ 2
= 162 cm²
A(EFGH) = (2×JH)² ÷ 2
= (2×4)² ÷ 2
= 32 cm²
Therefore:
A = A(ABCD) - <span>A(EFGH)
= 162 - 32
= 130 cm</span>²
The area of the shaded region is 130 cm².