4. Suppose the daily commute to campus for a professor has an approximately norm

Question

4. Suppose the daily commute to campus for a professor has an approximately normal distribution with mean 24 minutes and 3.8 minutes.
a. What is the probability for a randomly selected day the professor’s commute is under 20 minutes?
b. What is the probability for a randomly selected day the professor’s commute is over 35 minutes?
c. What is the probability for a randomly selected day the professor’s commute is between 25 and 30 minutes?
d. How long are the longest 10% of the professor’s commutes?

e. Suppose 10 days are chosen at random and that the commute time for any given day is independent of any other day.
i. What is the probability the average commute over the 10 days is under 20 minutes?
ii. What is the probability the average commute over the 10 days is between 25 and 30 minutes?
iii. How many of the 10 days would you expect to have a commute time under 20 minutes?
iv. What’s the probability fewer than 2 of these days have a commute time under 20 minutes?
v. What’s the probability exactly 2 of these days have a commute time under 20 minutes?

Thank you so much for any help received!

Explanation / Answer

Q.4 Commute time for professor = 24 minutes

Standard deviation of commute time for professor = 3.8 minutes

a. Lets say commute time is X and X follows normal distribution

Pr(X < 20 minutes) = NORM (X < 20; 24 ; 3.8)

Z= (20 - 24)/ 3.8 = -1.05

so Pr(X < 20) = (-1.05)

where is the standard cumulative normal distribution.

from Z - table , we fin

Pr(X < 20) = 0.1469

(b)

Pr(X > 35 minutes) = NORM (X > 35; 24 ; 3.8) = 1 - NORM (X =< 35; 24 ; 3.8)

Z= (35 - 24)/ 3.8 = 2.89

so Pr(X > 35) = 1 - (2.89)

where is the standard cumulative normal distribution.

from Z - table , we fin

Pr(X > 35) = 1 -  0.9981 = 0.0019

(c) Pr(25 minutes < X > 30 minutes) = NORM (X < 30; 24 ; 3.8) - NORM (X =< 25; 24 ; 3.8)

Z - values

Z2 = (30 - 24)/ 3.8 = 1.58 ; Z1 = (25 - 24)/ 3.8 = 0.26

Pr(25 minutes < X > 30 minutes) = (1.58) - (0.26)

= 0.9429 - 0.6141 = 0.3288

(iv) Lets say above X0 there are longest commute for professor

Pr( X > X0 ) = NORM (X > X0 ; 24 ; 3.8) = 0.1

NORM (X >  X0 ; 24 ; 3.8) = 1 - NORM (X <  X0 ; 24 ; 3.8)

NORM (X <  X0 ; 24 ; 3.8) = 1 - 0.9 = 0.1

so Z - value for the p - value = 0.1 from Z - table

Z = 1.645

so (X0 - 24)/ 3.8 = 1.645

X0 = 24 + 1.645 * 3.8 = 30.25 minutes

e. Sample size n = 10

Let say average commute over 10 days = x

Standard error of the average commute time se = / sqrt(n) = 3.8 / sqrt(10) = 1.20

(i) Pr(x < 20) = NORM (x < 20, 24 ; 1.20)

Z = (20 - 24)/1.2 = -3.33

Pr(x < 20) = (-3.33) = 0.0004

(ii) Pr(25 mins < x < 30 minutes) = NORM(25 minutes < x; 24; 1.20) - NORM( 30 minutes < x ; 24 ; 1.20)

Z2 = (25 - 24)/1.2 = 0.83, Z1 = (30 - 24)/1.2 = 5

Pr(25 mins < x < 30 minutes) = (5) - (0.83) = 0.9999 - 0.7967 = 0.2032

(iii) as Pr(X < 20 minutes) = 0.1469

so out of 10 days there are 10 * 0.1469 = 1.47 ~1.5 days when commute time under 20 minutes.

(iv) Pr(t <= 2; 10 ; 0.1469) = BIN(t <=2 ; 10 ; 0.146) (now it is the binomial distribution)

= 0.8305

(iv) Pr(t = 2; 10 ; 0.1469) = BIN(t =2 ; 10 ; 0.146) = 0.2714

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