Respuesta :
Using the normal distribution, it is found that there is a 0.8888 = 88.88% probability that his tube last the whole trip.
In a normal distribution 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]
- It 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 X.
- For n instances of a normal variable, the mean is [tex]n\mu[/tex] and the standard deviation is [tex]s = \sigma\sqrt{n}[/tex]
In this problem:
- For each time he brushes his teeth, the mean is of 0.13 ounces and the standard deviation is of 0.02 ounces, hence [tex]\mu = 0.13, \sigma = 0.2[/tex]
- He will brush his teeth 6 times, hence [tex]n = 6, n\mu = 6(0.13) = 0.79, s = 0.2\sqrt{6}[/tex]
The probability that his tube last the whole trip is the p-value of Z when X = 0.85, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
Considering the n instances
[tex]Z = \frac{X - n\mu}{s}[/tex]
[tex]Z = \frac{0.85 - 0.79}{0.2\sqrt{6}}[/tex]
[tex]Z = 1.22[/tex]
[tex]Z = 1.22[/tex] has a p-value of 0.8888.
0.8888 = 88.88% probability that his tube last the whole trip.
A similar problem is given at https://brainly.com/question/24663213
