Question

McMillan Assembly has a contract to assemble components for radar systems to be used by the U.S. military. The time required to complete one part of the assembly can be described by a Normal distribution with a mean of 29 hours and a standard deviation of 5 hours.

(a) What is the IQR of these assembly times? (4 decimals)

(b) The production manager would like to investigate the longest 5% of assembly times to see why it took so long for them to be assembled. At which assembly time should the production manager start investigating? (4 decimals)

Answer 1

The IQR of the assembly times is approximately -0.745. The production manager should start investigating at an assembly time of approximately 20.775 hours.

Step-by-step explanation:

(a) To find the IQR (Interquartile Range), we need to find the first quartile and the third quartile. The first quartile is the value below which 25% of the data falls, and the third quartile is the value below which 75% of the data falls. Using the z-score formula, we can calculate the z-scores for the first and third quartiles: Q1 = Mean - (z * Standard Deviation) and Q3 = Mean + (z * Standard Deviation), where z is the z-score corresponding to the desired percentile. For Q1, the desired percentile is 25%, so the z-score is -0.6745. Plugging in the values, we get Q1 = 29 - (-0.6745 * 5) = 32.3725. For Q3, the desired percentile is 75%, so the z-score is 0.6745. Plugging in the values, we get Q3 = 29 + (0.6745 * 5) = 31.6275. The IQR is calculated by subtracting Q1 from Q3: IQR = Q3 - Q1 = 31.6275 - 32.3725 = -0.745.

(b) To find the assembly time at which the production manager should start investigating, we need to find the z-score corresponding to the desired percentile. The desired percentile is 5%, so the z-score is -1.645. Plugging in the values, we get X = Mean + (z * Standard Deviation) = 29 + (-1.645 * 5) = 20.775.