Final answer:
To find the probability that the sample mean strength of company A will be at least 600kg more than that of company B, we compare the difference in sample means to 600kg using the standard error of the difference. The probability is approximately 0.
Step-by-step explanation:
To find the probability that the sample mean strength of company A will be at least 600kg more than that of company B, we need to find the difference in sample means and compare it to 600kg. The difference in sample means is the difference between the mean strengths of the two companies, which is 4500kg - 4000kg = 500kg. The standard error of the difference in sample means can be calculated using the formula:
standard error = sqrt((standard deviation of A)^2/nA + (standard deviation of B)^2/nB)
where nA is the number of wires from company A (50) and nB is the number of wires from company B (100). Plugging in the values, we get:
standard error = sqrt((200kg)^2/50 + (300kg)^2/100) = sqrt(80 + 90) = sqrt(170) ≈ 13.04kg
The z-score can then be calculated as:
z = (difference in sample means - 600kg)/(standard error) = (500kg - 600kg)/(13.04kg) ≈ -7.66
Using a z-table or calculator, we can find the probability associated with a z-score of -7.66, which is virtually 0. So the probability that the sample mean strength of company A will be at least 600kg more than that of company B is approximately 0.