Question

6.143 Cigarette Smoke Exposure. Researchers G. Evans and E. Kantrowitz explored the health consequences for exposure to many different environmental risks in the journal article "Socioe- conomic Status and Health: The Potential Role of Environmental Risk Exposure" (Annual Review of Public Health, Vol. 23, No. 1, pp. 303-331). According to research. 65% of preschool children living in poverty have been exposed to cigarette smoke at home. In comparison, 45% of preschool children not in poverty have been exposed to cigarette smoke at home. a. If 200 preschool children living in poverty are selected at ran- dom, what is the probability that at least 125 have been ex- posed to cigarette smoke at home? b. Repeat part (a) for children not living in poverty. 6.144 Naturalization. The U.S. Citizenship and Immigra tion Services collects and reports information about naturalized persons in Staristical Yearbook. During one year, there were 463.204 persons who became naturalized citizens of the United States and, of those, 41.5% were originally from some country in Asia. If 200 people who became naturalized citizens of the United States that year are selected at random, what is the probability that the number who were originally from some Asian country is a. fewer than 75 b. between 80 and 90, inclusive? e. less than 70 or more than 90? 6.145 High School Graduates. According to the document Current Population Survey, published by the U.S. Census Bureau, 31.6% of U.S. adults 25 years old or older have a high school de- gree as their highest educational level. If 100 such adults are se- lected at random, determine the probability that the number who have a high school degree as their highest educational level is a. exactly 32 b. between 30 and 35. inclusive. c. at least 25.

Answer 1

Sol:

6.144)

Let X denotes the number of people who became naturalized citizens out of 200 randomly selected people.

X ~ Binomial(n = 200, p = 0.415)

The probability mass function of X is

200 P(X = z) = *0.415"*(1 – 0.415 200-1, 1 = 0,1,2, ..., 200

Now,

a)

200 P(X < 75) = P(X < 74) * 0.415" * (1 – 0.415) 200-r 0.110829

b)

P(80 < X < 90) = P(X <90) - P(X < 79) = 0.858988 – 0.308864 = 0.550124

c)

P(X <70) + P(X >90) = P(X < 69) + [1 – P(X < 89)

→ P(X <70) + P(X >90) = 0.0255155+ (1 - 0.824671)

—P(X <70) + P(X >90) = 0.200844

6.145)

Let X denotes the number of US adults 25 years old or older have a high school degree as their highest education level out 100 adults.

X ~ Binomial(n = 100, p = 0.316)

The probability mass function of X is

P(X = z) = *0.316"* (1 -0.316)100-, 1 = 0,1,2, ..., 100

a)

100 P(X =32) = | *0.31632 *(1 – 0.316) 100-32 = 0.0849524 32

b)

P(30 < X < 35) = P(X < 35) - P(X < 29) = 0.800307 – 0.329688 = 0.470619

c)

P(x > 25) = 1- P(x < 25) = 1 - P(X < 24) = 1 – 0.0608037 = 0.939196

6.149)

p =0.40 , n = 250

a)

p = 0.40
P(P < 0.40)
z =( P - p ) / sqrt(npq)
= ( 0.40 - 0.40)/sqrt(250*0.0.44 * 0.56)
= -0.0161

P(P < 0.40) = P(z <-0.0161) = 0.4936 by using standard normal table

b)
P(x > 0.40)
z =( P - p ) / sqrt(npq)
= ( 0.40 - 0.40)/sqrt(250*0.0.40 * 0.56)
= -0.0161

P(P > 0.40) = P(z >-0.0161) = 0.5064 by using standard normal table

c)
mean = np = 250 * 0.40 = 100
p(x<90)
z =(x - mean ) / sqrt(npq)
= ( 90 - 100)/sqrt(250*0.40 * 0.56)
= -2.5482
p(x<90) = P(z < -2.5482) = 0.0054 by using standard normal table

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