Probability that workday bus ride exceeds weekend bus ride Duration of a bus ride from my house to the office any day of the work-week follows normal distribution N(30,10). A bus ride during the weekend, same route, home-office, follows N(20,5). What is the probability that the duration of the bus ride on Tuesday exceeds the bus ride on Saturday?
We can say that a bus ride on any single day is an independent event and the distribution is the same during the day, Monday-Friday (we are not accounting for a rush hour or a bus delay), or Saturday-Sunday.
UPD: in N(30,10), 30 is the mean and 10 is the variance. Hope this clears the confusion.
Hope this makes sense. Thank you!
 A: I will outline the method. 
(1) On Tuesday your travel time $T$ is
normally distributed with mean $\mu_T = 30,$ variance
$\sigma_T^2 = 10,$ and standard deviation $\sigma_T = 3.162.$
(2) On Saturday your travel time $S$ is
normally distributed with mean $\mu_S = 20,$ variance
$\sigma_S^2 = 5,$ and standard deviation $\sigma_S = 2.236.$
(3) As @BrianTung hinted, the difference $D$ in travel times is normally distributed with mean $\mu_D = \mu_T - \mu_S = 30 - 20 = 10.$ You may have this fact in your text or notes as something like: $E(X \pm Y) = E(X) \pm E(Y).$
Because you say Tuesday and Saturday travel times are independent, the variance of the difference is
$\sigma_D^2 = \sigma_T^2 + \sigma_S^2 = 10 + 5 = 15.$
[Be careful here: Notice that independence in required and that the variances are added even though the random variables are subtracted. Also, notices that variances add, but standard deviations do not.] 
Be sure to look in your text or notes for an equation that supports this adding of variances. My crystal ball says there is a 78.3% chance something like this will show up on an axam.
Thus $D$ is a normal random variable with $\mu_D = 10$ and $\sigma_D^2 = 15.$ I will leave it to you to use
standard methods to find $P(D > 0).$

Here is a histogram of a sample of a million hypothetical travel time distances $D = T - S.$
The density curve is for the normal distribution
derived above. 

The R code for the simulation is shown below. 
Notice that R uses the standard deviation $\sigma$ as the second parameter.] Technically, d > 0 is a
'logical' vector consisting of a million TRUEs & FALSEs, its mean is the proportion of its TRUEs.
With a million iterations, the approximation of
$P(D > 0)$ should be accurate to about three decimal
places.
set.seed(412)   # for reproducibility
m = 10^6
t = rnorm(m, 30, sqrt(10))
s = rnorm(m, 20, sqrt(5))
d = t - s
mean(d);  var(d)
[1] 9.99688     # aprx E(D) = 10
[1] 15.00828    # aprx Var(D) = 15
mean(d > 0);  mean(t > s)
[1] 0.995112    # aprx P(D > 0) = ??
[1] 0.995112    # another method for same

