It has been determined that Cindy\'s commuting time to work is normally distribu

Question

It has been determined that Cindy's commuting time to work is normally distributed with a mean is 17.2 and a standard deviation of 6.2 minutes, while her commuting time home normally distributed with a mean of and standard deviation of 7.3 minutes. Assume that these distributions are independent. The com times are recorded for a particular day. a) Find the probability that the sum of the commuting times is less than 40 minutes. b) Find the probability that the commute to work takes longer than the commute home.

Explanation / Answer

Q.4 Lets say X is the commuting Time to work of Cindy's and Y is the commuting time to Home of Cindy 's. Both are normal in nature

so X ~ Normal (17.2, 6.2)

Variance of X = 6.22

Y ~ Normal ( 19.6 , 7.3)

Variance of Y = 7.32

Lets say Z = X +Y = total time in commuting

so mean of Z = E(X + Y) = E(X) + E(Y) = 17.2 + 19.6 = 36.8 minutes

Variance of Z = 6.22 + 7.32 = 91.73

Std. dev.(Z) = 9.5775

(a) To find that total commuting time is less than 40 minutes

so Pr( Z <= 40; 36.8 ; 9.5775) = ?

z - value = ( 40- 36.8 )/ 9.5775 = 0.3342

so Pr( Z <= 40; 36.8 ; 9.5775) = (0.3342) = 0.6308

(b) Lets say W = X-Y

so Mean of W = E(W) = E(X) - E(Y) = 17.2 - 19.6 = -2.4 minutes

standard deviation will be the same as of Z

so S(Z) = 9.5775

so We have to calculate that

Pr( W > 0; -2.4, 9.5775 ) = 1- P(W<0;  -2.4, 9.5775)

so z = (2.4 -0) / 9.5775 = 0.2506

so P - value from Z - table

Pr( W > 0; -2.4, 9.5775 ) = 1 - (0.2506) = 1 - 0.5987 = 0.4013

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