Μ = (0×0.026) + (1×0.072) +(2×0.152) + (3×0.303) + (4×0.215) + (5×0.164) + (6×0.066)
μ = 0 + 0.072 + 0.304 + 0.909 + 0.86 + 0.82 + 0.396
μ = 3.361 ≈ 3.4
We need the value of ∑X² to work out the variance
∑X² = (0²×0.026) + (1²×0.072) + (2²×0.152) + (3²×0.303) + (4²×0.215) + (5²×0.164) + (6²×0.066)
∑X² = 0+0.072+0.608+2.727+3.44+4.1+2.376
∑X² = 13.323
Variance = ∑X² - μ²
Variance = 13.323 - (3.4)² = 1.763 ≈ 2
Standard Deviation = √Variance = √1.8 = 1.3416... ≈ 1.4
The correct answer related to the value of mean and standard deviation is the option D
<span>
An employee works an average of 3.4 overtime hours per week with a standard deviation of approximately 1.4 hours.</span>
G(0)= 0-1= -1
f(-1)= 3(-1)+5= -3+5= 2
the answer to the question is 2
Answer:
We have the system:
x ≤ 7
x ≥ a
Now we want to find the possible values of a such that the system has, at least, one solution.
First, we should look at the value of a where the system has only one solution:
We can write the 2 sets as:
a ≤ x
x ≥ 7
So, writing both together:
a ≤ x ≤ 7
if a is larger than 7, we do not have solutions.
then a = 7 gives:
7 ≤ x ≤ 7
Here the only solution is 7.
Now, if a is smaller than 7, for example 5, we have:
5 ≤ x ≤ 7
Now x can take different values, so we have a lot of solutions.
Then the restrictions for a, such that the system has at least one solution, is:
a ≤ 7.
