The length of time to find a parking space at 9 A.M. follows a normal distributi

Question

The length of time to find a parking space at 9 A.M. follows a normal distribution with a moan of 4 minutes and a standard deviation of 2 minutes. Eighty percent of the time, it takes more than how many minutes to find a parking space? min The length of time to find a parking space at 9 A.M. follow, a normal distribution with a mean of 6 minutes and a standard deviation of 2minutes. Based upon the above information would it be unusual if it took less than 1 minute to find a parking space? Yes No unable to determine

Explanation / Answer

Given mean µ =4

Standard deviation =2

Given probability =0.8

Let us consider more than x min to fins the parking space of probability 0.8 is

P(X >x) =0.8

i.e. P(X < x) =0.2

P(Z < x-µ /) = 0.2

so x-µ / =invNorm(0.2)

(x-4)/2 = -0.8416                     [-from excel formula =NORMSINV(0.2) =-0.84162123]

x = 2.32

2nd part

Given mean µ =6

Standard deviation =2

Probability of less than 1 minutes P(X <1)

P(X-µ / < x-µ /) =P(z< x-µ /)

                                        =P(z < 1-6 / 2)

                                        =P(z < -2.5)

                                        =0.006        

[from excel formula =NORMSDIST(-2.5) =0.00620967]

The probability 0.0062 < 0.5, So YES it is unusual it took less than 1 minute to find a parking space

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