banner

The amount of time Ricardo spends brushing his teeth follows a normal distribution with unknown mean and standard deviation. Ricardo spends less than one minute brushing his teeth about 40% of the time. He spends more than two minutes brushing his teeth 2% of the time. Use this information to determine the mean and standard deviation of this distribution.

 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:

\[\sigma=0.44\]

5/5 - (18 votes)