Question 1194732: Each minute a machine produces a length of rope with
mean of 4 feet and standard deviation of 5 inches. Assuming that the
amounts produced in different minutes are independent and
identically distributed, approximate the probability that the machine will produce at least
250 feet in one hour. (Use the central limit theorem, ∅ = (0.26) 0.6026
)
Answer by ElectricPavlov(122)   (Show Source):
You can put this solution on YOUR website! **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.
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