Normal distribtion (a little confused?) On a particular day, 50% of the employees arrive at work by 8.30 am, and 10% had not arrived by 8.55 am. 
Assuming a normal model, find the standard deviation of the arrival times, in minutes?
What I tried: From what I understand, 40% of the people showed up in the gap of 25 minutes. Therefore, taking the difference between 0.90 - 0.50 = 0.40.
My query is how will I form an equation that I can solve because I don't have a x-bar, mean and standard deviation (3 unknowns?)
 A: The normal probability distribution has several nice properties that can help resolve your confusion. Consider in particular its symmetry: half of the cases have values lower than the mean of the distribution (the median equals the mean). So you do have information about the mean of the distribution of arrival times. You also have information about the 90th percentile of arrival times, when 10% hadn't yet arrived. That's enough information to estimate the standard deviation based on normal probability tables.
A: As median = mean for the normal distribution you know that: $$\mu\hat{=}\text{8:30}$$. 
Now the question is how to deal with the time Data on a numeric scale. But if we only want to find out the sd in minutes we could simply define that 8:30 corresponds to $830$. And e.g. 9:30 would be $830+60=890$ as 9:30 is $60$ minutes later. But we could equally well set 8:30 to $0$ and $9:30$ to $60$ (the difference is still measured in minutes), this change would not alter the sd.
Furthermore you know (using 8:30 corresponds to $830$) that $$\Phi_{830,\sigma ^2}(855)=0.9$$ (denoting with $\Phi_{\mu,\sigma ^2}$ the distribution function of a $N(\mu, \sigma ^2)$ 
This is equivalent to 
$$\Phi_{0,1}(\frac{855-830}{\sigma})=0.9$$
Now you only have to apply the inverse oh the distribution function of the $N(0,1)$ and solve for $\sigma$
