Questions: A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 165.1 cm and a standard deviation of 0.8 cm. For shipment, 12 steel rods are bundled together. Round all answers to four decimal places if necessary. a. What is the distribution of X ? X ~ N( ) b. What is the distribution of x̄ ? x̄ ~ N( ) c. For a single randomly selected steel rod, find the probability that the length is between 165 cm and 165.2 cm.

Solution Steps

a. The distribution of $X$ is given as $N(\mu, \sigma)$, where $\mu$ is the mean and $\sigma$ is the standard deviation. Here, $\mu = 165.1$ cm and $\sigma = 0.8$ cm.

b. The distribution of $\bar{x}$ (the sample mean) for a sample size $n$ is also normally distributed with mean $\mu$ and standard deviation $\sigma / \sqrt{n}$. Here, $n = 12$.

c. To find the probability that a single randomly selected steel rod has a length between 165 cm and 165.2 cm, we need to calculate the cumulative distribution function (CDF) for a normal distribution with the given mean and standard deviation.

The lengths of the steel rods are normally distributed with a mean $\mu = 165.1$ cm and a standard deviation $\sigma = 0.8$ cm. Therefore, the distribution of $X$ is: $$X \sim N(165.1, 0.8)$$

For a sample size $n = 12$, the distribution of the sample mean $\bar{x}$ is also normally distributed with the same mean $\mu = 165.1$ cm and a standard deviation of $\frac{\sigma}{\sqrt{n}}$. Thus, the standard deviation of $\bar{x}$ is: $$\sigma_{\bar{x}} = \frac{0.8}{\sqrt{12}} \approx 0.2309$$ Therefore, the distribution of $\bar{x}$ is: $$\bar{x} \sim N(165.1, 0.2309)$$

To find the probability that a single randomly selected steel rod has a length between 165 cm and 165.2 cm, we use the cumulative distribution function (CDF) for a normal distribution with mean $\mu = 165.1$ and standard deviation $\sigma = 0.8$.

The probability is given by: $$P(165 \leq X \leq 165.2) = \Phi\left(\frac{165.2 - 165.1}{0.8}\right) - \Phi\left(\frac{165 - 165.1}{0.8}\right)$$ Using the CDF values, we get: $$P(165 \leq X \leq 165.2) \approx 0.0995$$

Final Answer