Unformatted text preview: Student: Grady Simonton Course: Math119: Elementary Statistics - Spring 2010 - CRN: 49239
Instructor: Shawn Par'vini - 16 weeks Date: 3/18/10 Book: Triola: Elementary Statistics, 11e
Time: 4:07 PM Women's heights are normally distributed with mean 63.4 in and standard deviation of 2.5 in. A social organization for
tall people, has a requirement that women must be at least 70 in tall. What percentage of women meet that requirement? Suppose that a random variable x is normally distributed with mean u and standard deviation 6. The area below the
normal curve represents a proportion or probability. To help visualize the area of interest, draw a standard normal curve. Label the given values
for x and n. Then shade the region representing the desired probability. Since we are
interested in women taller than 70 in, shade the area to the right of 70. 63.4 70
. . . X — l1
The next step rs to convert 70 to its corresponding z-score. Use the fact that z = o .
70 — 63.4 2 64
z = — = .
2.5 You can either use a standard normal distribution table or technology to ﬁnd the area under the normal curve. For this
explanation, a standard normal distribution table is used. The standard normal distribution table gives the cumulative area under the normal curve to the left of the z-score. Recall
that the area under the normal curve is equal to 1. To ﬁnd the area under the normal curve to the right of the z-score,
subtract the cumulative area to the left of the z-seore from 1. Use a standard normal distribution table to ﬁnd the cumulative area to the left of z = 2.64. P(z < 2.64) = 0.9959 Now subtract the area 0.9959 from 1 to ﬁnd the area of the shaded region. Shaded area = l — 0.9959 = 0.0041. To ﬁnd percentage of women that are taller than 70 in, convert 0.0041 to a percent. 0.0041 =0.41% Page 1 ...
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- Spring '10