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.

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.

Transcript text: A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of $165.1-\mathrm{cm}$ and a standard deviation of $0.8-\mathrm{cm}$. For shipment, 12 steel rods are bundled together. Round all answers to four decimal places if necessary. a. What is the distribution of $X$ ? $X \sim \mathrm{N}($ $\square$ $\square$ b. What is the distribution of $\bar{x} ? \bar{x}-\mathrm{N}($ $\square$ $\square$ c. For a single randomly selected steel rod, find the probability that the length is between $165-\mathrm{cm}$ and $165.2-\mathrm{cm}$. $\square$

Solution

Solution Steps

Solution Approach

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.

Step 1: Determine the Distribution of $ X $

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: \[ X \sim N(165.1, 0.8) \]

Step 2: Determine the Distribution of $ \bar{x} $

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: \[ \bar{x} \sim N(165.1, 0.2309) \]

Step 3: Calculate the Probability for a Single Steel Rod

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

The distribution of $ X $ is $ X \sim N(165.1, 0.8) $.
The distribution of $ \bar{x} $ is $ \bar{x} \sim N(165.1, 0.2309) $.
The probability that a single steel rod is between 165 cm and 165.2 cm is $ \boxed{0.0995} $.

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