Let Xi be the random variable representing the number of units the first worker produces in day i.
Define X = X1 + X2 + X3 + X4 + X5 as the random variable representing the number of units the
first worker produces during the entire week. It is easy to prove that X is normally distributed with mean µx = 5·75 = 375 and standard deviation σx = 20√5.
Similarly, define random variables Y1, Y2,...,Y5 representing the number of units produces by
the second worker during each of the five days and define Y = Y1 + Y2 + Y3 + Y4 + Y5. Again, Y is normally distributed with mean µy = 5·65 = 325 and standard deviation σy = 25√5. Of course, we assume that X and Y are independent. The problem asks for P(X > Y ) or in other words for P(X −Y > 0). It is a quite surprising fact that the random variable U = X−Y , the difference between X and Y , is also normally distributed with mean µU = µx−µy = 375−325 = 50 and standard deviation σU, where σ2 U = σ2 x+σ2 y = 400·5+625·5 = 1025·5 = 5125. It follows that σU = √5125. A reference to the above fact can be found online at http://mathworld.wolfram.com/NormalDifferenceDistribution.html.
Now everything reduces to finding P(U > 0) P(U > 0) = P(U −50 √5125 > − 50 √5125)≈ P(Z > −0.69843) ≈ 0.757546 .
Answer:
A- m= 686 - 65h
B- h = 686 + 65m
C- m = 686 + 65h
D- h = 686 - 65m
step-by-step explanation:
When ever I encounter problem like this, I make the equation become:
z = anything buy z.
First, you want to get 2/5 to the other side, to do so, you need to multiply 5/2 on both side. then you are left with z+1=5/2 y. Then you subtract 1 from both sides to make z on its own. The answer should be z = 3/2y or z= 1
y
2/5(z+1)=y
2/5(z+1)5/2=5/2y
z+1=5/2y
z+1-1=5/2y-1
z=3/2y or z=<span>1
y</span>
15 is your percentage to the answer