Use your calculator to determine the probabilities. State what you put in for th

Question

Use your calculator to determine the probabilities. State what you put in for the LOWER LIMIT, UPPER LIMIT, MEAN and STANDARD DEVIATION.

Maple tree diameters in a forest area are normally distributed with mean 10 inches and standard deviation 2.2 inches. Find the PERCENTAGE of trees having a diameter greater than 15 inches

2. White blood cell (WBC) count per cubic millimeter of whole blood follows approximately a Normal distribution with mean 7500 and standard deviation 1750. What percentage of people have WBC between 7000 and 8000?

3. Lifetimes of a certain brand of tires is approximately normally distributed with mean 42,500 miles and standard deviation 3,200 miles. What percentage of tires will last more than 50,000 miles?

4. The incomes of a set of factory workers happen to be normally distributed. The average income is $53,000 and the standard deviation is $9,000 What is the probability that a randomly selected employee makes more than $65,000? What is the probability that the average of 4 randomly selected employees makes more than $65,000 What is the probability that the average of 12 randomly selected employees makes more than $65,000?

Explanation / Answer

In TI83 CALC

Step 1: Press the 2nd key and then press VARS then 2 to get “normalcdf.”

Step 2: Enter the following numbers into the screen:
99999 for the lower bound, followed by a comma, then 15for the upper bound, followed by another comma.

Step 3: Press 10(for the mean), followed by a comma and then 2.2(for the standard deviation).

Step 4: Close the argument list with a “)”. (Your display should now read normalcdf(15, 99999, 10, 2.2).)

answer:

.0115*100=1.15%

SOlution2:

normalcdf(7000, 8000, 7500, 1750).)

=0.2249

0.2249*100=22.49%

Solution3:

normalcdf(50000, 99999, 42500, 3200).)

=0.0095

=0.0095*100=0.95%

Solution4:

hat is the probability that a randomly selected employee makes more than $65,000?

normalcdf(65000, 99999, 53000, 9000).)

=0.0912

=0.0912*100=9.12%

the probability that the average of 4 randomly selected employees makes more than $65,000

std error=9000/sqrt(4)=9000/2=4500

normalcdf(65000, 99999, 53000, 4500).)

=0.0038

=0.0038*100=0.38%

the probability that the average of 12 randomly selected employees makes more than $65,000

stderror=9000/sqrt(12)

=2598.076

normalcdf(65000, 99999, 53000, 2598.076).)

=1.93*10-6

=1.93*10-6*100

=0.0000193

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