Question 1194732
**1. Define Variables**

* Let X be the length of rope produced in one minute.
* Let S be the total length of rope produced in one hour (60 minutes).

**2. Given**

* Mean length of rope per minute (μ_X) = 4 feet
* Standard deviation of rope per minute (σ_X) = 5 inches = 5/12 feet 
* Number of minutes in an hour (n) = 60

**3. Apply Central Limit Theorem**

* Since S is the sum of 60 independent and identically distributed random variables (X), the Central Limit Theorem states that the distribution of S will be approximately normal.

* **Mean of Total Length (μ_S):**
    * μ_S = n * μ_X = 60 minutes * 4 feet/minute = 240 feet

* **Standard Deviation of Total Length (σ_S):**
    * σ_S = √(n) * σ_X = √(60) * (5/12) feet ≈ 3.229 feet

**4. Standardize the Value**

* We want to find P(S ≥ 250 feet)
* Standardize 250 feet:
    * z = (X - μ_S) / σ_S 
    * z = (250 feet - 240 feet) / 3.229 feet 
    * z ≈ 3.09

**5. Find the Probability**

* Using a standard normal distribution table or a calculator:
    * P(S ≥ 250) = P(Z ≥ 3.09) ≈ 0.001

**Therefore, the approximate probability that the machine will produce at least 250 feet of rope in one hour is 0.001 (or 0.1%).**

**Key Points:**

* The Central Limit Theorem allows us to approximate the distribution of the sum of a large number of independent and identically distributed random variables as normal, even if the individual variables themselves are not normally distributed.
* In this case, we assumed that the length of rope produced in each minute is independent of the length produced in other minutes.
* The calculation involves standardizing the value of interest (250 feet) and then using the standard normal distribution to find the corresponding probability.