To complete the first setup on a new machine took an employee 300 minutes. using an 88???% cumulative average minus time learnin
1 answer:
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Answer:
A) 0.99413
B) 0.00022
Step-by-step explanation:
A) First of all let's find the total grading time from 6:50 P.M to 11:00 P.M.:
Total grading time; X = 11:00 - 6:50 = 4hours 10minutes = 250 minutes
Now since we are given an expected value of 5 minutes, the mean grading time for the whole population would be:
μ = n*μ_s ample = 42 × 5 = 210 minutes
While the standard deviation for the population would be:
σ = √nσ_sample = √(42 × 6) = 15.8745 minutes
To find the z-score, we will use the formula;
z = (x - μ)/σ
Thus;
z = (250 - 210)/15.8745
z = 2.52
From the z-distribution table attached, we have;
P(Z < 2.52) ≈ 0.99413
B) solving this is almost the same as in A above, the only difference is an additional 10 minutes to the time.
Thus, total time is now 250 + 10 = 260 minutes
Similar to the z-formula in A above, we have;
z = (260 - 210)/15.8745
z = 3.15
P(Z > 3.15) = 0.00022
