Question 1166864
The probability that the mean weight of your sample falls within a certain number of standard deviations from the population mean can be determined using the **Empirical Rule** (or 68-95-99.7 Rule) for a Normal distribution.

The problem implies that the distribution of the **sample mean** ($\bar{X}$) is also Normal (due to the original distribution being Normal), and we are measuring distances in terms of the standard deviation of the sample mean (the **Standard Error**, $\sigma_{\bar{X}}$).

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## A. Within 2 Standard Deviations from the Mean

The probability that the sample mean is within **2 standard deviations** of the population mean ($\mu \pm 2\sigma_{\bar{X}}$) is approximately **95%**.

* $P(\mu - 2\sigma_{\bar{X}} < \bar{X} < \mu + 2\sigma_{\bar{X}}) \approx \mathbf{0.95}$ or $\mathbf{95\%}$

This value comes directly from the Empirical Rule, which states that approximately 95% of the data in a Normal distribution lies within two standard deviations of the mean. 

[Image of a normal distribution curve with the area within 2 standard deviations shaded]


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## B. More than 1 Standard Deviation from the Mean

The probability that the sample mean is **more than 1 standard deviation** from the mean is the area in the two tails combined (i.e., outside the range $\mu \pm 1\sigma_{\bar{X}}$).

1.  **Find the area within 1 standard deviation:**
    According to the Empirical Rule, the probability that the sample mean is **within 1 standard deviation** of the population mean is approximately **68%**.
    $$P(\mu - 1\sigma_{\bar{X}} < \bar{X} < \mu + 1\sigma_{\bar{X}}) \approx 0.68$$

2.  **Calculate the area outside 1 standard deviation:**
    Since the total area under the probability curve is $1$ (or $100\%$), the probability of the sample mean being *more than* $1$ standard deviation away is $1$ minus the probability of it being *within* $1$ standard deviation.

    $$\text{Probability (More than 1 SD away)} = 1 - P(\text{Within 1 SD})$$
    $$\text{Probability (More than 1 SD away)} \approx 1 - 0.68$$
    $$\text{Probability (More than 1 SD away)} \approx \mathbf{0.32}$$ or $\mathbf{32\%}$

* $P(|\bar{X} - \mu| > 1\sigma_{\bar{X}}) \approx \mathbf{0.32}$ or $\mathbf{32\%}$