A system starts working at time t = 0. Its lifetime has approximately a normaldistribution with mean value μ = 120 and standard deviation σ = 24 *hours+. After afailure, the system is replaced by an equivalent new one in negligible time andimmediately resumes its work. How many spare systems must be available in order tobe able maintain the replacement process over an interval of length 1000 hours(1) with probability 0.60?(2) with probability 0.79 ?

To determine the number of spare systems needed to maintain the replacement process over a 1000-hour interval with a given probability, we can use the concept of the normal distribution.

Step 1: Calculate the z-score The first step is to calculate the z-score corresponding to the desired probability. The z-score is calculated using the formula:

z = (x - μ) / σ

where x is the value we want to find the probability for, μ is the mean, and σ is the standard deviation.

For part (1) with a probability of 0.60, we need to find the z-score corresponding to the 60th percentile. Using a standard normal distribution table or a calculator, we can find that the z-score for a 60th percentile is approximately 0.253.

For part (2) with a probability of 0.79, we need to find the z-score corresponding to the 79th percentile. Using a standard normal distribution table or a calculator, we can find that the z-score for a 79th percentile is approximately 0.765.

Step 2: Calculate the number of spare systems Once we have the z-score, we can use it to calculate the number of spare systems needed. The formula to calculate the number of spare systems is:

n = z * σ / (x - μ)

where n is the number of spare systems, z is the z-score, σ is the standard deviation, and x is the length of the interval (1000 hours in this case).

For part (1): n = 0.253 * 24 / (1000 - 120) n ≈ 0.006 * 24 / 880 n ≈ 0.00015 * 24 n ≈ 0.0036

Since we cannot have a fraction of a system, we need at least 1 spare system.

For part (2): n = 0.765 * 24 / (1000 - 120) n ≈ 0.01836 * 24 / 880 n ≈ 0.44064

Since we cannot have a fraction of a system, we need at least 1 spare system.

Therefore, for part (1) with a probability of 0.60, we need at least 1 spare system. And for part (2) with a probability of 0.79, we also need at least 1 spare system.