Respuesta :
Let [tex]X[/tex] denote the number of minutes spent brushing his teeth, and let [tex]\mu[/tex] be the mean and [tex]\sigma[/tex] the standard deviation for this distribution.
[tex]\mathbb P(X<1)=\mathbb P\left(\dfrac{X-\mu}\sigma<\dfrac{1-\mu}\sigma\right)=0.40[/tex]
The z-score corresponding to this probability is approximately [tex]z=-0.2533[/tex], which means
[tex]\dfrac{1-\mu}\sigma=-0.2533\iff\mu-0.2533\sigma=1[/tex]
Next, (note the sign change)
[tex]\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[/tex]
The corresponding z-score is approximately [tex]z=2.0538[/tex], so you have
[tex]\dfrac{2-\mu}\sigma=2.0538\iff\mu+2.0538\sigma=2[/tex]
Solving the two equations for [tex]\mu[/tex] and [tex]\sigma[/tex], you'll find that the mean is approximately [tex]\mu=1.1098[/tex] and the standard deviation is approximately [tex]\sigma=0.4334[/tex].
[tex]\mathbb P(X<1)=\mathbb P\left(\dfrac{X-\mu}\sigma<\dfrac{1-\mu}\sigma\right)=0.40[/tex]
The z-score corresponding to this probability is approximately [tex]z=-0.2533[/tex], which means
[tex]\dfrac{1-\mu}\sigma=-0.2533\iff\mu-0.2533\sigma=1[/tex]
Next, (note the sign change)
[tex]\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[/tex]
The corresponding z-score is approximately [tex]z=2.0538[/tex], so you have
[tex]\dfrac{2-\mu}\sigma=2.0538\iff\mu+2.0538\sigma=2[/tex]
Solving the two equations for [tex]\mu[/tex] and [tex]\sigma[/tex], you'll find that the mean is approximately [tex]\mu=1.1098[/tex] and the standard deviation is approximately [tex]\sigma=0.4334[/tex].
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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
- 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:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]-0.253 = \frac{1 - \mu}{\sigma}[/tex]
[tex]1 - \mu = -0.253\sigma[/tex]
[tex]\mu = 1 + 0.253\sigma[/tex]
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:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]2.054 = \frac{2 - \mu}{\sigma}[/tex]
[tex]2 - \mu = 2.054\sigma[/tex]
[tex]\mu = 2 - 2.054\sigma[/tex]
Since [tex]\mu = 1 + 0.253\sigma[/tex]
[tex]1 + 0.253\sigma = 2 - 2.054\sigma[/tex]
[tex]2.307\sigma = 1[/tex]
[tex]\sigma = \frac{1}{2.307}[/tex]
[tex]\sigma = 0.43[/tex]
And [tex]\mu = 1 + 0.253(0.43) = 1.11[/tex]
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
