group 2000 patients, means clot time for blood was 8sec with std deviation of 3sec.Times r normally distrib?

What per centage of the patients had a blood-clotting time between 8 and 14 seconds?
a. 68% B. 34% c. 49.5% d. 47.5%

How many of the patients have a blood-clotting time between 5 and 11 seconds?

a. 68 b 34 c 1360 d 680

What is the probability that a patient from this group had a clotting time more than 11 seconds?

a .025 b .16 c .34 d.135

Assuming that the patient’s blood clot times are normally distributed, we can use the normal cdf function on a calculator to calculate the probabilities

The general form of the normal cdf function is

normalcdf(left end point, right end point, mean, std deviation)

a)

So on TI83, TI84, TI89, etc, simply type in to get

normalcdf(8,14, 8, 3) = 0.47

So the answer is d.

b)

Use the normalcdf function to find the percentage of people that had blood clot times between 5 and 11 seconds:

normalcdf(5, 11, 8, 3) = 0.68

So 68% of the population had blood clot times between 5 seconds and 11 seconds

Now multiply 0.68 by 2000 to get

0.68*2000=1360

So there are 1,360 people who had blood clot times between 5 seconds and 11 seconds

c)

Now use the normal cdf function to find the area under the curve from 11 to infinity

normalcdf(11, E99, 8, 3) = 0.1586

note: E99 is a very large number that is used in place of infinity

So the probability that a patient from this group had a clotting time more than 11 seconds is 16% which means that the answer is b.

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