### Statistics Problem: Jiffy Peanut Butter Jars#### Problem Statement3. Jiffy Peanut Butter jars (labeled as a 32 oz. jar) have an average weight of 32.2 oz with σ = 0.25 oz. A random sample of 36 jars is taken:a) Verify conditions of the sample mean weight of the 36 jars being normally distributed and then state **mean** and **standard deviation** for the sample mean.b) What is the probability that the sample mean weight of 36 jars is between 32.15 and 32.22 ounces?---### Solution Guide#### Part (a)**Normal Distribution Verification:**The conditions for the sample mean to be normally distributed are generally met if the sample size is sufficiently large, typically n ≥ 30, owing to the Central Limit Theorem. Given n = 36, this condition is satisfied.**Mean and Standard Deviation of the Sample Mean:**- **Mean (μ):** The mean of the sample mean is the same as the population mean. Therefore, μ = 32.2 oz. - **Standard Deviation (σ_x̄):** The standard deviation of the sample mean (σ_x̄) is the population standard deviation (σ) divided by the square root of the sample size (n).\[ \sigma_{x̄} = \frac{\sigma}{\sqrt{n}} = \frac{0.25}{\sqrt{36}} = \frac{0.25}{6} = 0.0417 \, \text{oz} \]#### Part (b)**Probability Calculation:**We need to find the probability that the sample mean weight (x̄) is between 32.15 oz and 32.22 oz.Firstly, standardize the values to find the z-scores:\[ z = \frac{x̄ - μ}{σ_{x̄}} \]For x̄ = 32.15 oz:\[ z_{32.15} = \frac{32.15 - 32.2}{0.0417} = -1.2 \]For x̄ = 32.22 oz:\[ z_{32.22} = \frac{32.22 - 32.2}{0.0417} = 0.48 \]Using the standard normal distribution table (or a calculator):

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3. Jiffy Peanut Butter jars (labeled as a 32 oz. jar) have average weight of 32.2 oz with o 0.25 oz. A random sample of 36 jars is taken: a) Verify conditions of the sample mean weight of the 36 jars being normally distributed and then state mean and standard deviation for the sample mean. b) What is the probability that the sample mean weight of 36 jars is between 32.15 and 32.22 ounces?

3. Jiffy Peanut Butter jars (labeled as a 32 oz. jar) have average weight of 32.2 oz with o 0.25 oz. A random sample of 36 jars is taken: a) Verify conditions of the sample mean weight of the 36 jars being normally distributed and then state mean and standard deviation for the sample mean. b) What is the probability that the sample mean weight of 36 jars is between 32.15 and 32.22 ounces?

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Estimation

A new framework of data analysis where the new statistical methods are used for interpreting the results and analyzing the data is known as estimation in statistics.

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### Statistics Problem: Jiffy Peanut Butter Jars

#### Problem Statement

3. Jiffy Peanut Butter jars (labeled as a 32 oz. jar) have an average weight of 32.2 oz with σ = 0.25 oz. A random sample of 36 jars is taken:

a) Verify conditions of the sample mean weight of the 36 jars being normally distributed and then state **mean** and **standard deviation** for the sample mean.

b) What is the probability that the sample mean weight of 36 jars is between 32.15 and 32.22 ounces?

---

### Solution Guide

#### Part (a)
**Normal Distribution Verification:**

The conditions for the sample mean to be normally distributed are generally met if the sample size is sufficiently large, typically n ≥ 30, owing to the Central Limit Theorem. Given n = 36, this condition is satisfied.

**Mean and Standard Deviation of the Sample Mean:**

- **Mean (μ):** The mean of the sample mean is the same as the population mean. Therefore, μ = 32.2 oz.

- **Standard Deviation (σ_x̄):** The standard deviation of the sample mean (σ_x̄) is the population standard deviation (σ) divided by the square root of the sample size (n).

\[ \sigma_{x̄} = \frac{\sigma}{\sqrt{n}} = \frac{0.25}{\sqrt{36}} = \frac{0.25}{6} = 0.0417 \, \text{oz} \]

#### Part (b)
**Probability Calculation:**

We need to find the probability that the sample mean weight (x̄) is between 32.15 oz and 32.22 oz.

Firstly, standardize the values to find the z-scores:

\[ z = \frac{x̄ - μ}{σ_{x̄}} \]

For x̄ = 32.15 oz:

\[ z_{32.15} = \frac{32.15 - 32.2}{0.0417} = -1.2 \]

For x̄ = 32.22 oz:

\[ z_{32.22} = \frac{32.22 - 32.2}{0.0417} = 0.48 \]

Using the standard normal distribution table (or a calculator):

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