Respuesta :
Answer:
The travel time is at least 28 minute.
Step-by-step explanation:
Given : Abby tracks her travel time to work for 60 days and determines that her mean travel time, in minutes, is 35.6. assume the travel times are normally distributed with a standard deviation of 10.3 min.
To determine : The travel time x such that 22.96% of the 60 days have a travel time that is at least x.
Solution :
Mean, μ=35.6 min
Standard deviation, σ=10.3 min
we are required to find the value of x such that 22.96% of the 60 days have a travel time that is at least x.
Using z-table, the z-score that will give us 0.2296 is -0.74 (Shown in attached graph)
The formula of z-score is given by:
[tex]Z-score=\frac{x-\mu}{\sigma}[/tex]
[tex]-0.74=\frac{x-35.6}{10.3}[/tex]
[tex]-0.74\times 10.3=x-35.6[/tex]
[tex]-7.622=x-35.6[/tex]
[tex]-7.622+35.6=x[/tex]
[tex]x=27.978[/tex]
Approximately the value of x is 28.
The travel time is at least 28 minute.
Using the normal distribution, it is found that x = 43.2 minutes.
---------------------
Normal Probability Distribution
Problems of normal distributions can be solved using the z-score formula, which in a set mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], is given by, for a measure X:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
- The Z-score measures how many standard deviations the measure is from the mean.
- Each z-score has a p-value, which is the probability that the value of the measure is smaller than X, that is, the percentile of X.
- The probability that the measure is greater than X is 1 subtracted by the p-value.
---------------------
- Mean of 35.6 means that [tex]\mu = 35.6[/tex]
- Standard deviation of 10.3 means that [tex]\sigma = 10.3[/tex].
- The value x is found when Z has a p-value of 1 - 0.2296 = 0.7704, thus X when Z = 0.74. Then:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]0.74 = \frac{X - 35.6}{10.3}[/tex]
[tex]X - 35.6 = 10.3(0.74)[/tex]
[tex]X = 43.2[/tex]
Thus, x = 43.2 minutes.
A similar problem is given at https://brainly.com/question/15061321
