Monitors manufactured by tsi electronics have life spans that have a normal distribution with a variance of 1,000,0001,000,000 and a mean life span of 19,00019,000 hours. if a monitor is selected at random, find the probability that the life span of the monitor will be more than 19,80019,800 hours. round your answer to four decimal places.

The probability is 0.4861.

The probability that X>19,800 is the probability that the z-score is greater than the given z-score.  We must compute that first:



However, we know the variance, not the standard deviation.  The standard deviation is the square root of the variance:

√1000000 = 1000

Now we have:


Using a z-table (http://www.z-table.com) we see that the probabilities given are to the left of the z-score, which means less than.  We want P(z>0.8).  Thus we will compute 1-P(z<0.8) = 1-0.5319=0.4681.

I’m not sure what sujevtbis this