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horrorfan [7]
4 years ago
11

9) A construction company employs 2 sales engineers. Engineer 1 does the work in estimating cost for 70% of jobs bid by the comp

any. Engineer 2 does the work for 30% of jobs bid by the company. It is known that the error rate for Engineer 1 is such that 0.02 is the probability of an error when he does the work, whereas the probability of an error in the work of Engineer 2 is 0.04. What is the probability that a bid provided by the company will contain an error in the cost estimates?
If a serious error is known occurred, what is the probability that it was made by Engineer 2?
Engineering
1 answer:
Irina18 [472]4 years ago
3 0

The probability that the error occurred when Engineer 2 made the mistake is 0.462 on the other hand the probability that the error occurred when the engineer 1 made the mistake is 0.538

Explanation:

Let E_{1} denote the event that the 1st  engineer  does the work.so we write

P(E)_{1}=0.7

Let E_{2} denote the event that the 2nd engineer  does the work .So we write

P(E)_{2}=0.3

Let O denote the event during which the error occurred .so we write

P(O/E_{1} )=0.02(GIVEN)

P(O/E_{2} )=0.04(GIVEN)

  • The probability that the error occurred when the first engineer performed the work is P(E_{1} /O)
  • The probability that the error occurred when the first engineer performed the work is P(E_{2} /O)

Now we need to find when did the error in the work occur so we will compare the probability of the work done by <u>engineer 1 </u>and <u>engineer 2 </u>

<u></u>

<u>lets find the Probability of the Engineer 1</u>

<u>Using Bayes theorem,we get</u>

<u></u>

<u></u>P(E_{1} /O) =0.02*0.7/0.02*0.7+0.04*0.3 = 0.014/0.026=0.538

<u>lets find the Probability of the Engineer 2</u>

<u></u>P(E_{2} /O) =0.04*0.3/0.02*0.7+0.04*0.3=0.012/0.026=0.462

Since ,0.462<0.538 so it is more prominent that the Engineer 1 did the work when the error occurred

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Answer:

the range of K can be said to be :  -3.59 < K< 0.35

Explanation:

The transfer function of a typical tape-drive system is given by;

KG(s) = \dfrac{K(s+4)}{s[s+0.5)(s+1)(s^2+0.4s+4)]}

calculating the characteristics equation; we have:

1 + KG(s) = 0

1+   \dfrac{K(s+4)}{s[s+0.5)(s+1)(s^2+0.4s+4)]} = 0

{s[s+0.5)(s+1)(s^2+0.4s+4)]} +{K(s+4)}= 0

s^5 + 1.9 s^4+ 5.1s^3+6.2s^2+ 2s+K(s+4) = 0

s^5 + 1.9 s^4+ 5.1s^3+6.2s^2+ (2+K)s+ 4K = 0

We can compute a Simulation Table for the Routh–Hurwitz stability criterion Table as  follows:

S^5             1                     5.1                          2+ K

S^4            1.9                   6.2                           4K

S^3             1.83            \dfrac{1.9 (2+K)-4K}{1.9}          0

S^2        \dfrac{11.34-1.9(X)}{1.83}       4K                         0

S          \dfrac{XY-7.32 \ K}{Y}        0                            0

\dfrac{1.9 (2+K)-4K}{1.9} = X

 

\dfrac{11.34-1.9(X)}{1.83}= Y

We need to understand that in a given stable system; all the elements in the first column is usually greater than zero

So;

11.34  - 1.9(X) > 0

11.34  - 1.9(\dfrac{3.8+1.9K-4K}{1.9}) > 0

11.34  - (3.8 - 2.1K)>0

7.54 +2.1 K > 0

2.1 K > - 7.54

K > - 7.54/2.1

K > - 3.59

Also

4K >0

K > 0/4

K > 0

Similarly;

XY - 7.32 K > 0

(\dfrac{3.8+1.9K-4K}{1.9})[11.34  - 1.9(\dfrac{3.8+1.9K-4K}{1.83}) > 7.32 \ K]

0.54(2.1K+7.54)>7.32 K

11.45 K < 4.07

K < 4.07/11.45

K < 0.35

Thus the range of K can be said to be :  -3.59 < K< 0.35

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