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 zscore. 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 zscore. 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 zscore,
subtract the cumulative area to the left of the zseore 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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