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# 6.5.q7 - Student Grady Simonton Course Math119 Elementary...

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Unformatted text preview: Student: Grady Simonton Course: Math119: Elementary Statistics - Spring 2010 - CRN: 49239 Instructor: Shawn Parvini - 16 weeks Date: 3/18/10 Book: Triola: Elementary Statistics, 11e Time: 4:14 PM An engineer is going to redesign an ejection seat for an airplane. The seat was designed for pilots weighing between 150 lb and 201 lb. The new population of pilots has normally distributed weights with a mean of 156 1b and a standard deviation of 34.5 lb. a. If a pilot is randomly selected, ﬁnd the probability that his weight is between 150 lb and 201 lb. Notice that an individual value from a normally distributed population has been chosen. Therefore, use the population distribution to determine the probability. x-u First, convert the given weights to their corresponding 2 scores using 2 = and the population distribution statistics. 150- 156 34.5 =5 —0.17 20] - 156 34.5 as 1.3 The probability is the area between 2 = - 0.17 and z = 1.3 under the standard normal distribution. §=150 x=201 u;=156 Find the area by subtracting the area to the left of z = — 0.1? from the area to the left of z = 1.3. Use a table to ﬁnd each of these areas. The area to the left ofz = - 0.17 is 0.4325. The area to the left ofz = 1.3 is 0.9032. Subtract to ﬁnd the area between 2 = — 0.17 and z = 1.3. 0.9032 — 0.4325 = 0.470? Therefore, the probability that a randomly selected pilot's weight is between 150 lb and 201 lb is approximately 0.4707. b. If 32 different pilots are randomly selected, ﬁnd the probability that their mean weight is between 150 lb and 201 lb. In this case, the desired probability is for the mean of a sample of 32 pilots. Therefore, use the central limit theorem. According to the central limit theorem, the distribution of sample means i will have a mean given by 1.1;: p and a Page 1 Student: Grady Simonton Course: Math119: Elementary Statistics - Spring 2.010 - CRN: 492.39 Instructor: Shawn Parvini - 16 weeks Date: 3/18/10 Book: Triola: Elementary Statistics, 11e Time: 4:14 PM 0 standard deviation given by c; = T 11 The mean of the distribution of sample means ; is the same as the population mean, so it; = 156. Apply the deﬁnition for the standard deviation of the distribution of the sample means for a sample size of 32. 0' O— x n 34.5 W Simplify to ﬁnd the standard deviation of the distribution of the sample means. Lo) m4}- N'U. >4l =66. CD 98796 Therefore, the distribution of sample means i for a sample size of 32 is approximately normal with a mean it; = 156 and a standard deviation 6; = 6.098796. Use these values to compute the corresponding 2 score for each height. 150 - 156 6.098796 =5 — 0.98 2‘35 201 - 156 6.098796 =5 7.38 2‘35 The probability is the area between 2 = - 0.98 and z = 7.38 under the standard normal distribution. M I II Ix.) o ._. §=150 n;=156 Find the area by subtracting the area to the left of z = - 0.98 from the area to the left of z = 7.38. Use a table to ﬁnd each of these areas. The area to the left ofz = — 0.98 is 0.1635. The area to the left ofz = 7.38 is 0.9999. Page 2. Student: Grady Simonton Course: Math119: Elementary Statistics - Spring 2.010 - CRN: 49239 Instructor: Shawn Parvini - 16 weeks Date: 3/18/10 Book: Triola: Elementary Statistics, 11e Time: 4:14 PM Subtract to ﬁnd the area between 2 = — 0.98 and z = 7.38. 0.9999 — 0.1635 = 0.8364 Therefore, the probability that the mean weight of a sample of 32 randomly selected pilots is between 150 lb and 201 lb is approximately 0.8364. c. When redesigning the ejection seat, which probability is more relevant? If it is more important how the seat behaves for an individual, then the result from part (a) is more relevant. If it is more important how the seat behaves on average for a group of people, then the result from part (b) is more important. Page 3 ...
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