Statistics Problem Need Help!

Question

The time it takes a symphony orchestra to play Beethovens Ninth Symphony has a normal distribution with a mean of 64.3 minutes and a standard deviation of 1.15 minutes. The next time it is played what is the probability that it will take between 62.5 and 67.7 minutes?  

Answer 1

You need to calculate two z-scores, one for each of the values 62.5 and 67.7, using the formula z = (x-μ)/σ.

 

z1 = (62.5-64.3)/1.15 = -1.56

z2 = (67.7-64.3)/1.15 = 2.96

 

Convert each z-score to a probability (also called a p-value) using a probability table or calculator.

 

P(Z<-1.56) = 0.05938

P(Z<2.96) =  0.99846

 

Subtract the smaller from the larger:

0.99846 - 0.05938 = 0.93908

 

We can express this probability as a percentage by moving the decimal point two places to the right...

so 0.939 becomes 93.9 percent.

 

There is a 93.9 percent probability that the time it takes the orchestra to play Beethoven's Ninth Symphony will be between 62.5 minutes and 67.7 minutes.

 

Notice that the p-values we calculated tell us more about what's happening. The p-value 0.05938 (associated with 62.5 minutes) means that we can expect the orchestra to play the symphony in less than 62.5 minutes only 5.9 percent of the time. The p-value 0.99846 (associated with 67.7 minutes) means that we can expect the orchestra to play the symphony in more than 67.7 minutes less than one percent of the time because 1 - 0.99846 = .00154 which is .154 percent.