SOLUTION: The ticket sales for events held at the new civic center are believed to be normally distributed with a mean of 12,000 and a standard deviation of 1,000. a. What is the prob

Question 51698: The ticket sales for events held at the new civic center are believed to be normally distributed with a mean of 12,000 and a standard deviation of 1,000.
a. What is the probability of selling more than 10,000 tickets?
b. What is the probability of selling between 9,500 and 11,000 tickets?
c. What is the probability of selling more than 13,500 tickets?

Found 2 solutions by stanbon, mikeebsc:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Let me know if you still need help with these problems.
Cheers,
Stan H.
Answer by mikeebsc(26) (Show Source): You can put this solution on YOUR website!
A. We must find the z-score first, so we use the formula:

( X - MEAN)/Standard Deviation.
X is the given variable 10000, so we have:
(10000 - 12000)/1000 = -2.
First, we drop the negative sign in the answer to this equation because X < MEAN. We then look for the value that matches a z-score of 2.000 in the standard normal distribution table and find it is .9772.
So P>10000=.9772
TI-83 and TI-84 Calculator steps for those that are allowed to use one vs using a table:
2nd
DISTR
Scroll down to #2: normalcdf
**the upper boundary is positive infinity, which we can represent with 1,000,000
**the lower boundary is given(10,000) since you are looking for "more than" this number
normalcdf(lowerbound, upperbound, mean, standard deviation) **or**
normalcdf(10000,1000000,12000,1000)
Press enter twice and you will get .9772, the same as using the standard normal distribution table. If asked to convert to a %, just multiply by 100 and add a % sign at the end. Probability and percent chance are the same thing.

B. Again we must first find the z-scores for both variables (9500 and 11000), so we use the same formula:
(X-MEAN)/Standard Deviation
(9500 - 12000)/1000 = -2.5
(11000 - 12000)/1000 = -1
Then we look at the standard normal distribution table and find the corresponding z scores (again changing the negative symbols because the X values are < MEAN):
z-score for 2.500 = .9938
z score for 1.000 = .8413
Now we subtract the z-scores: .9938 - .8413 = .1525
So P=>9500 and P<=11000 = .1525
**For TI-83 and TI-84 users**
Repeat the above calculator steps using 9500 as the lowerbound and 11000 as the upperbound

C. Again we must find the z-score first using the formula:
(X-MEAN)/Standard Deviation **or**
(13500 - 12000)/1000 = 1.5
Now we make this a -1.5 (add a negative symbol) because the X >MEAN (greater than) and find the corresponding z-score in the standard normal distribution table which is .0668 so
P>13500 = .0668
**Using a TI-83 or TI-84**
Repeat the above steps listed in problem A with 13500 as the lowerbound and 1000000 as the upperbound as you are asked to find more than 13500, so this makes it the lowest boundary.

**Yes there is a better way to explain when you drop or add the negative symbols, but I would need access to a video feed or whiteboard to show you as it involves drawing a bell curve. Suffice it to say that the rules I gave you that if x < mean drop the - symbol or x > mean add a negative symbol to look it up in the standard normal distribution table is the absolute easiest way to explain it. Always check your work and just look at it after you are done to see if your answer "makes sense". Good luck and I hope this helps!!

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