Suppose the weekly amount of money spent on maintenance and repairs by a company is NOT approximately normally distributed. Howe
1 answer:
Answer:
Probability = 0.00621
Step-by-step explanation:
We are given;
Population mean; μ = 400
Population standard deviation; σ = 20
Raw score; x¯ = 450
the probability that the actual costs will exceed the budgeted amount will be the p-value.
Thus, let's first find the z-score after which we will find the p-value.
z = (x¯ - μ)/σ
z = (450 - 400)/20
z = 50/20
z = 2.5
From online p-value from z-score calculator attached, using z = 2.5; one tailed hypothesis; significance level of 0.05, we have;
P-value = 0.00621
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Answer:
For n = 3, there are 134 possibilities
Step-by-step explanation:
First, lets calculate the generating function.
For each square we have 2 possibilities: red and blue. The Possibilities between n² squares multiply one with each other, giving you a total of 2^n² possibilities to fill the chessboard with the colors blue or red.
However, rotations are to be considered, then we should divide the result by 4, because there are 4 ways to flip the chessboard (including not moving it), that means that each configuration is equivalent to three other ones, so we are counting each configuration 4 times, with the exception of configurations that doesnt change with rotations.
A chessboard that doesnt change with rotations should have, in each position different from the center, the same colors than the other three positions it could be rotated into. As a result, we can define a <em>symmetric by rotations chessboard</em> with only (n²-1)/4 + 1 squares (the quarter part of the total of squares excluding the center plus the center).
We cocnlude that the total of configurations of symmetrical boards is
Since we have to divide by 4 the rest of configurations (because we are counted 4 times each one considering rotations), then the total number of configutations is
If n = 3, then the total amount of possibilities are