Problem 4 The time to microwave a bag of popcorn using the

Problem # 4 The time to microwave a bag of popcorn using the automatic setting can be treated as a random variable having a normal distribution standard deviation 10 secohds. If the probability is 0.8212 that the bag will take less than 282.5 seconds to pop, find the probability that it will take longer than 258.3 seconds to pop

Solution

For a left tailed area of 0.8212, the corresponding z score is

z = 0.919947873

As

u = x-z*sigma

Then

u = 282.5 - 0.919947873*10 = 273.3005213

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    258.3      
u = mean =    273.3005213      
          
s = standard deviation =    10      
          
Thus,          
          
z = (x - u) / s =    -1.50005213      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   -1.50005213   ) =    0.93319955 [ANSWER]