Mathematics, 01.09.2019 01:10 imbadatmath56

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

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Answer from: Dweath50

Using the information and solving a system of equations, it is found that:

The mean of the distribution is of 1.11 minutes.The standard deviation is of 0.43 minutes.

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Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The z-score measures how many standard deviations the measure is from the mean. After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of measure X.

Ricardo spends less than one minute brushing his teeth about 40% of the time.

This means that Z when X = 1 has a p-value of 0.4, thus, when X = 1, Z = -0.253. Then:

$Z = \frac{X - \mu}{\sigma}$

$-0.253 = \frac{1 - \mu}{\sigma}$

$1 - \mu = -0.253\sigma$

$\mu = 1 + 0.253\sigma$

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:

$Z = \frac{X - \mu}{\sigma}$

$2.054 = \frac{2 - \mu}{\sigma}$

$2 - \mu = 2.054\sigma$

$\mu = 2 - 2.054\sigma$

Since $\mu = 1 + 0.253\sigma$

$1 + 0.253\sigma = 2 - 2.054\sigma$

$2.307\sigma = 1$

$\sigma = \frac{1}{2.307}$

$\sigma = 0.43$

And $\mu = 1 + 0.253(0.43) = 1.11$

The mean of the distribution is of 1.11 minutes.

The standard deviation is of 0.43 minutes.

A similar problem is given at link

Answer from: camhappy2005

Let

denote the number of minutes spent brushing his teeth, and let $\mu$ be the mean and $\sigma$ the standard deviation for this distribution.

$\mathbb P(X$

The z-score corresponding to this probability is approximately z=-0.2533

, which means

$\dfrac{1-\mu}\sigma=-0.2533\iff\mu-0.2533\sigma=1$

Next, (note the sign change)

$\mathbb P(X2)=0.02\implies\mathbb P(X\le2)=\mathbb P\left(\dfrac{X-\mu}\sigma\le\dfrac{2-\mu}\sigma\right)=0.98$

The corresponding z-score is approximately z=2.0538

, so you have

$\dfrac{2-\mu}\sigma=2.0538\iff\mu+2.0538\sigma=2$

Solving the two equations for $\mu$ and $\sigma$ , you'll find that the mean is approximately $\mu=1.1098$ and the standard deviation is approximately $\sigma=0.4334$ .

Answer from: Quest

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5would be the answer- explanation

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