Answer:
b is the answer I believe so because it have to be greater than 0
Step-by-step explanation:
I hope it's right and if not my bad
Answer:
4
Step-by-step explanation:
p(3) = 2³ = 8
p(2) = 2² = 4
8-4 = 4
Answer:
(A)20%, 20%, 20%, 20%, 20%
Step-by-step explanation:
We are told that 20% of the M&M's produced by the Mars Corporation are Orange in color. If 5 students take a random sample of 50 M&M's, the most plausible percentage of orange candies that will be obtained is:
20%, 20%, 20%, 20%, 20%
This is as a result of the fact that in research experiments, a sample is a representation of the whole. Any result which holds for a random sample should hold for the total population and vice versa.
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 .
