The **question aims** to find the mean $\mu$ and standard deviation $\sigma$ of a **standard normal distribution. **

In arithmetic, a **standard score** is the number of standard deviations where the maturity of the observed point is above or below the average value of what is observed or measured. **Raw scores** above the mean generally have **positive points,** while those with less than the mean have** negative scores. Standard scores** are often called **z-scores**; both terms can be used interchangeably. Other equivalent words include** z values,** **common points, and variables**.

## Expert Answer

**Common distribution** problems can be solved using the **z-score formula.** In a set with **mean** $\mu$ and **standard deviation** $\sigma$, the** z-value** of the scale X is given:

\[Z=\dfrac{X-\mu}{\sigma}\]

- $Z$-score measures how many
**standard deviations**are derived from the description. - After
**finding**the $z-score$, we look at the**z-score**table and find the $p-value$ associated with that $z-score$, which is the $X$**percentage point.**

**Ricardo spends less than one minute brushing his teeth** about $40\%$ of the time. The time is **more than two minutes** about $2\%$ of the time, and thus **less than two minutes** about $98\%$of the time.

The $z-value$ is **calculated** by:

This** means** that $Z$ When $X=1$ has a $p-value$ of $0.4$, thus when $X=1$, $Z=-0.253$ then:

\[Z=\dfrac{X-\mu}{\sigma}\]

\[-0.253=\dfrac{1-\mu}{\sigma}\]

\[1-\mu=-0.253\sigma\]

\[\mu=1+0.253\sigma\]

He spends more than two minutes brushing his teeth $2\%$ of the time. This means that $Z$ when $X = 2$ has a $p-value$ of $1 – 0.02 = 0.98$, thus, when $X = 2$,$ Z = 2.054$, then:

\[Z=\dfrac{X-\mu}{\sigma}\]

\[2.054=\dfrac{2-\mu}{\sigma}\]

\[2-\mu=2.054\sigma\]

\[\mu=2-2.054\sigma\]

**Since,**

\[\mu=1+0.253\sigma\]

\[(1+0.253\sigma)=(2-2.054\sigma)\]

\[2.307\sigma=1\]

\[\sigma=0.43\]

**The value** of the $\sigma$ is $0.43$.

**The value** of the $\mu$ is calculated as:

\[\mu=1+0.253(0.43)\]

\[\mu=1.11\]

**The value** of the $\mu$ is $1.11$.

## Numerical Results

The **value of the mean** $\mu$ is **calculated** as:

\[\mu=1.11\]

The** value of the standard deviation** $\sigma$ is **calculated** as:

\[\sigma=0.43\]

## Example

**The time Bella spends brushing her teeth follows the normal distribution with an unknown definition and standard deviation. Bella spends less than one minute brushing her teeth about $30\%$ of the time. She spends more than two minutes brushing her teeth $4\%$ of the time. Use this information to find the mean and standard deviation from this distribution.**

**Solution**

**Bella spends less than one minute brushing her teeth** about $30\%$ of the time. The time is less than two minutes about $4\%$ of the time, and thus less than two minutes about $96\%$ of the time.

The $z-value$ is** calculated** by:

This **means** that $Z$ When $X=1$ has a $p-value$ of $0.3$, thus when $X=1$, $Z=-0.5244$ then:

\[Z=\dfrac{X-\mu}{\sigma}\]

\[-0.5244=\dfrac{1-\mu}{\sigma}\]

\[1-\mu=-0.5244\sigma\]

\[\mu=1+0.5244\sigma\]

She **spends more than two minutes brushing her teeth** 4% of the time. This means that $Z$ when $X = 2$ has a $p-value$ of $1 – 0.04 = 0.96$, thus, when $X = 2$,$ Z = 1.75069$. Then:

\[Z=\dfrac{X-\mu}{\sigma}\]

\[1.75069=\dfrac{2-\mu}{\sigma}\]

\[2-\mu=1.75069\sigma\]

\[\mu=2-1.75069\sigma\]

**Since,**

\[\mu=1+0.5244\sigma\]

\[(1+0.5244\sigma)=(2-1.75069\sigma)\]

\[2.27\sigma=1\]

\[\sigma=0.44\]

**The value** of the $\sigma$ is $0.44$.

**The value** of the $\mu$ is calculated as:

\[\mu=1+0.5244(0.44)\]

\[\mu=1.23\]

**The value of the mean** $\mu$ is calculated as:

\[\mu=1.23\]

**The value of the standard deviation** $\sigma$ is calculated as:

\[\sigma=0.44\]