First, we must ensure all units are consistent. Since the mean and the target question are in feet, we'll convert the standard deviation from inches to feet.
- Mean per minute $\left(\mu_X\right): 4$ feet
- Standard Deviation per minute $\left(\sigma_X\right): 5$ inches
- $\sigma_X($ in feet $)=\frac{5 \text { inches }}{12 \text { inches/foot }} \approx 0.4167$ feet
- Time ( n ): 1 hour $=60$ minutes
Apply the Central Limit Theorem (CLT)
The Central Limit Theorem states that the sum of a large number ( $n=60$ ) of independent and identically distributed random variables ( $X_i$ ) will be approximately normally distributed. Let $S_{60}$ be the total rope produced in 60 minutes.
We need to find the mean and standard deviation for this total sum ( $S_{60}$ ).
Mean of the Sum $\left(\mu_{S_{60}}\right)$ :
- $\mu_{S_{\text {601 }}}=n \times \mu_X$
- $\mu_{S_{\text {6il }}}=60 \times 4$ feet $=240$ feet
Standard Deviation of the Sum ( $\sigma_{S_{\text {Bid }}}$ ):
- $\sigma_{S_{\infty}}=\sqrt{n} \times \sigma_X$
- $\sigma_{S_{\infty}}=\sqrt{60} \times\left(\frac{5}{12}\right.$ feet $)$
- $\sigma_{S_{50}} \approx 3.2275$ feet
So, the total production in one hour is approximately normal with a mean of 240 feet and a standard deviation of 3.2275 feet.
We want to find the probability of producing at least 250 feet, which is $P\left(S_{60} \geq 250\right)$. To find this using the standard normal distribution, we calculate the Z-score:
- $Z=\frac{\text { Value-Mean }}{\text { Standard Deviation }}$
- $Z=\frac{250-240}{3.2275}$
- $Z \approx 3.0984$
We need to find the area under the standard normal curve to the right of our Z-score, which corresponds to $P(Z \geq 3.0984)$.
- $P(Z \geq 3.0984)=1-P(Z<3.0984)$
- Using a standard normal table or calculator, $P(Z<3.0984) \approx 0.999027$
- $P(Z \geq 3.0984)=1-0.999027$
- $P(Z \geq 3.0984) \approx \mathbf{0 . 0 0 0 9 7 3}$
