Answer: Given:
Mean (μ) = 12 months
Standard deviation (σ) = 2 months
We can use the standard normal distribution to find the probabilities.
a) Probability that an instrument produced by this machine will last less than 7 months.
We need to find P(X < 7), where X is the length of life of an instrument.
z-score = (7 - μ) / σ = (7 - 12) / 2 = -2.5
Using a standard normal distribution table or calculator, we can find that the probability of getting a z-score less than -2.5 is 0.0062.
Therefore, P(X < 7) = 0.0062 or approximately 0.62%.
b) Probability that an instrument produced by this machine will last between 7 and 12 months.
We need to find P(7 < X < 12).
z-score for X = 7: z1 = (7 - μ) / σ = (7 - 12) / 2 = -2.5
z-score for X = 12: z2 = (12 - μ) / σ = (12 - 12) / 2 = 0
Using a standard normal distribution table or calculator, we can find the area between z1 and z2:
P(-2.5 < Z < 0) = 0.4938
Therefore, P(7 < X < 12) = P(-2.5 < Z < 0) = 0.4938 or approximately 49.38%.
Explanation: