Finn changes his mind and, from now on, decides to take the normal route to work everyday. On any given day, the time (in minutes) it takes Finn to get to work is normally distributed with mean u = 35 and variance = 81 .Find the 33rd percentile of the time it takes Finn to get to work on any given day. Do not include any units in your answer.Recall from Question 6 that on any given day, the time (in minutes) it takes Finn to get to work is normally distributed with mean = 35 and variance =81. Assume the time it takes him to get to work is independent from day to day.Over the next 2 days, find the probability that Finn took more than 40.5 minutes to get to work on the first day or more than 38.5 minutes to get to work on the second day.

Answer:The 33rd percentile of the time it takes Finn to get to work on any given day is 31.04 minutes.There is a 61.92% probability that Finn took more than 40.5 minutes to get to work on the first day or more than 38.5 minutes to get to work on the second day.Step-by-step explanation:This can be solved by the the z-score formula:On a normaly distributed set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a value X is given by:[tex]Z = \frac{X - \mu}{\sigma}[/tex]Each z-score value has an equivalent p-value, that represents the percentile that the value X is:The problem states that:Mean = 35, so [tex]\mu = 35[/tex]Variance = 81. The standard deviation is the square root of the variance, so [tex]\sigma = \sqrt{81} = 9[/tex].Find the 33rd percentile of the time it takes Finn to get to work on any given day. Do not include any units in your answer.Looking at the z-score table, [tex]z = -0.44[/tex] has a pvalue of 0.333. So what is the value of X when [tex]z = -0.44[/tex].[tex]Z = \frac{X - \mu}{\sigma}[/tex][tex]-0.44 = \frac{X - 35}{9}[/tex][tex]X - 35 = -3.96[/tex][tex]X = 31.04[/tex]The 33rd percentile of the time it takes Finn to get to work on any given day is 31.04 minutes.Over the next 2 days, find the probability that Finn took more than 40.5 minutes to get to work on the first day or more than 38.5 minutes to get to work on the second day.[tex]P = P_{1} + P_{2}[/tex][tex]P_{1}[/tex] is the probability that Finn took more than 40.5 minutes to get to work on the first day. The first step to solve this problem is finding the z-value of [tex]X = 40.5[/tex].[tex]Z = \frac{X - \mu}{\sigma}[/tex][tex]Z = \frac{40.5 - 35}{9}[/tex][tex]Z = 0.61[/tex][tex]Z = 0.61[/tex] has a pvalue of 0.7291. This means that the probability that it took LESS than 40.5 minutes for Finn to get to work is 72.91%. The probability that it took more than 40.5 minutes if [tex]P_{1} = 100% - 72.91% = 27.09% = 0.2709[/tex][tex]P_{2}[/tex] is the probability that Finn took more than 38.5 minutes to get to work on the second day. Sine the probabilities are independent, we can solve it the same way we did for the first day, we find the z-score of [tex]X = 38.5[/tex][tex]Z = \frac{X - \mu}{\sigma}[/tex][tex]Z = \frac{38.5 - 35}{9}[/tex][tex]Z = 0.39[/tex][tex]Z = 0.39[/tex] has a pvalue of 0.6517. This means that the probability that it took LESS than 38.5 minutes for Finn to get to work is 65.17%. The probability that it took more than 38 minutes if [tex]P_{1} = 100% - 65.17% = 34.83% = 0.3483[/tex]So:[tex]P = P_{1} + P_{2} = 0.2709 + 0.3483 = 0.6192[/tex]There is a 61.92% probability that Finn took more than 40.5 minutes to get to work on the first day or more than 38.5 minutes to get to work on the second day.