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6.2.q15 - Student Grady Silnonton Course Math119 Elementary...

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Unformatted text preview: Student: Grady Silnonton Course: Math119: Elementary Statistics - Spring 2010 - CRN: 49239 Instructor: Shawn Parvini - 16 weeks Date: 3/18/10 Book: Triola: Elementary Statistics, 11e Time: 4:04 PM Assume the readings on thermometers are normally distributed with a mean of 0°C and a standard deviation of 100°C. Find the probability P( - 2.06 < z < 2.06), where z is the reading in degrees. In general, to find the area between 20 and 21 , use a standard normal distribution table to find the area to the left of 20 and to the left of z 1. The area between 20 and 21 is (area to the left of z 1) — (area to the left of 20). You can also use technology to find the area. To visualize the problem, draw a standard normal curve with z = - 2.06 and z = 2.06 labeled and the area of interest shaded. -2.06 2.05 The next step is to find the areas to the left of the z scores in question. The area to the left ofz = 2.06 is 0.9803. The area to the left ofz = - 2.06 is 0.0192. Now that you know the areas to the left of z = — 2.06 and z = 2.06, you can find the area between those values. Calculate the area between 2 = — 2.06 and z = 2.06. Area = (area left of z = 2.06) - (area left of z = — 2.06) = 0.9803 - 0.0197r = 0.9606 Therefore, P( — 2.06 < z 4 2.06) = 0.9606. Page 1 ...
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