**Find the mean difference in weights of the melons?****Find the standard deviation of the difference in weights?****If a Normal model can be used to describe the difference in weights, find the probability that the melon you got at the first store is heavier?**

This question aims to find the **mean difference** and **standard deviation** in the difference in **weights** of the **melons** from two stores. Also, to check if the melon from the **first** store is **heavier.**

The question is based on the concepts of **probability** from a **normal distribution** using a **z**-table or **z-score**. It also depends on the **population’s mean** and the **population’s standard deviation.** The **z-score** is the **deviation** of a data point from the **population’s mean.** The formula for **z-score** is given as:

\[ z = \dfrac{ x\ -\ \mu}{ \sigma } \]

## Expert Answer

The given information about this **problem** is as follows:

\[ Mean\ Weight\ of\ Melons\ from\ First\ Store\ \mu_1 = 22 \]

\[ Standard\ Deviation\ of\ Weight\ of\ Melons\ from\ First\ Store\ \sigma_1 = 2.5 \]

\[ Mean\ Weight\ of\ Melons\ from\ Second\ Store\ \mu_2 = 18 \]

\[ Standard\ Deviation\ of\ Weight\ of\ Melons\ from\ Second\ Store\ \sigma_2 = 2 \]

**a)** To calculate the **mean difference** between **weights** of the **melons** from the first and second store, we simply need to take the difference of the **means** of both stores. The **mean difference** is given as:

\[ \mu = \mu_1\ -\ \mu_2 \]

\[ \mu = 22\ -\ 18 \]

\[ \mu = 4 \]

**b)** To calculate the **standard deviation** in difference in **weights** of the **melons** from both stores, we can use the following formula which is given as:

\[ SD = \sqrt{ \sigma_1^2 + \sigma_2^2 } \]

Substituting the values, we get:

\[ SD = \sqrt{ 2.5^2 + 2^2 } \]

\[ SD = \sqrt{ 6.25 + 4 } \]

\[ SD = \sqrt{ 10.25 } \]

\[ SD = 3.2016 \]

**c)** The **normal model** of the differences in **mean** and **standard deviation** can be used to calculate the **probability** that the melon from the first store is **heavier** than the melon from the second store. The formula to calculate **z-score** is given as:

\[ z = \dfrac{ x\ -\ \mu}{ \sigma } \]

Substituting the values, we get:

\[ z = \dfrac{ 0\ -\ 4 }{ 3.2016 } \]

\[ z = -1.25 \]

Now we can calculate the **probability** using the z-table.

\[ P(Z \gt 1.25) = 1\ -\ P(Z \lt -1.25) \]

\[ P(Z \gt 1.25) = 1\ -\ 0.1056 \]

\[ P(Z \gt 1.25) = 0.8944 \]

## Numerical Result

**a)** The **mean difference** in the **weights** of the **melons** between the first and second store is calculated to be **4.**

**b)** The **standard deviation** of the **difference** in **weights** is calculated to be **3.2016.**

**c)** The **probability** that the **melon** from the **first** is **heavier** than the **melon** from the **second store** is calculated to be **0.8944 or 89.44%.**

## Example

The **mean** of a sample is given as **3.4** and the **standard deviation** of the sample is given as **0.3**. Find the **z-score** of a **random** sample of **2.9.**

The **formula** for **z-score** is given as:

\[ z = \dfrac{ x\ -\ \mu}{ \sigma } \]

Substituting the values, we get:

\[ z = \dfrac{ 2.9\ -\ 3.4 }{ 0.3 } \]

\[ z = -1.67 \]

The **probability** associated with this **z-score** is given as **95.25%.**