Answer:
B yes, because every x value corresponds to exactly one y value
Step-by-step explanation:
A function means that each input goes to a different output
Each input goes to only 1 y value
For large sample confidence intervals about the mean you have:
xBar ± z * sx / sqrt(n)
where xBar is the sample mean z is the zscore for having α% of the data in the tails, i.e., P( |Z| > z) = α sx is the sample standard deviation n is the sample size
We need only to concern ourselves with the error term of the CI, In order to find the sample size needed for a confidence interval of a given size.
z * sx / sqrt(n) = width.
so the z-score for the confidence interval of .98 is the value of z such that 0.01 is in each tail of the distribution. z = 2.326348
The equation we need to solve is:
z * sx / sqrt(n) = width
n = (z * sx / width) ^ 2.
n = ( 2.326348 * 6 / 3 ) ^ 2
n = 21.64758
Since n must be integer valued we need to take the ceiling of this solution.
n = 22
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>
