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
Let \(X\) denote the number of minutes spent brushing his teeth, and let \(\mu\) be the mean and \(\sigma\) the standard deviation for this distribution.
\(\mathbb P(X<1)=\mathbb P\left(\dfrac{X-\mu}\sigma<\dfrac{1-\mu}\sigma\right)=0.40\)
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)
\(\mathbb P(X>2)=0.02\implies\mathbb P(X\le2)=\mathbb P\left(\dfrac{X-\mu}\sigma\le\dfrac{2-\mu}\sigma\right)=0.98\)
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 \(\sigma=0.4334\).
Using the information and solving a system of equations, it is found that:
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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:
\(Z = \frac{X - \mu}{\sigma}\)
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 https://brainly.com/question/15061321
