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the length of time to complete a door assembly on an automobile factory assembly line is normally distributed with a mean of 6.7 minutes and a standard deviation of 2.2 minutes. For a door selected at random, what is the probability the assembly time will be: (a) 5 minutes or less? (b) 10 minutes or more? (c) between 5 and 10 minutes?

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To solve these probability questions, we'll use the properties of the normal distribution and the given mean (μ) and standard deviation (σ).

Given data:

Mean (μ) = 6.7 minutes

Standard deviation (σ) = 2.2 minutes

(a) Probability that the assembly time will be 5 minutes or less:

We need to find the area under the normal curve to the left of 5 minutes.

Z-score = (X - μ) / σ

Z-score for X = 5 minutes:

Z = (5 - 6.7) / 2.2 ≈ -0.773

Using a standard normal distribution table or calculator, we find the cumulative probability corresponding to Z = -0.773 is approximately 0.2206.

Therefore, the probability that the assembly time will be 5 minutes or less is approximately 0.2206 or 22.06%.

(b) Probability that the assembly time will be 10 minutes or more:

We need to find the area under the normal curve to the right of 10 minutes.

Z-score for X = 10 minutes:

Z = (10 - 6.7) / 2.2 ≈ 1.5

Using the standard normal distribution table or calculator, we find the cumulative probability corresponding to Z = 1.5 is approximately 0.9332.

To find the probability of 10 minutes or more, we subtract the cumulative probability from 1:

P(X ≥ 10) = 1 - 0.9332 ≈ 0.0668

Therefore, the probability that the assembly time will be 10 minutes or more is approximately 0.0668 or 6.68%.

(c) Probability that the assembly time will be between 5 and 10 minutes:

We need to find the area under the normal curve between 5 and 10 minutes.

Z-score for X = 5 minutes (from part a):

Z = -0.773

Z-score for X = 10 minutes (from part b):

Z = 1.5

Using a standard normal distribution table or calculator, we find the cumulative probabilities corresponding to Z = -0.773 and Z = 1.5:

P(X ≤ 5) ≈ 0.2206

P(X ≥ 10) ≈ 0.0668

To find the probability between 5 and 10 minutes, we subtract the cumulative probabilities:

P(5 < X < 10) = P(X ≥ 10) - P(X ≤ 5) ≈ 0.0668 - 0.2206 ≈ 0.1538

Therefore, the probability that the assembly time will be between 5 and 10 minutes is approximately 0.1538 or 15.38%.

User Tom Hallam
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