Suppose my iPod has 3000 songs The length of the songs has a

Suppose my iPod has 3000 songs. The length of the songs has a slightly right skewed distribution. The mean length is 3.45 minutes and standard deviation is 1.63 minutes. I’m about to take a 6 hours trip and I make a random playlist of 100 songs. What is the probability that my playlist last entire trip (Hint: That is equivalent of asking P(x >= 3.6)?

Solution

We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as          
          
x = critical value =    3.6      
u = mean =    3.45      
n = sample size =    100      
s = standard deviation =    1.63      
          
Thus,          
          
z = (x - u) * sqrt(n) / s =    0.920245399      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   0.920245399   ) =    0.178722268 [ANSWER]