7.14 In a recent survey of full-time female workers ages 22 to 35 years, 46% sai

Question

7.14 In a recent survey of full-time female workers ages 22 to 35 years, 46% said that they would rather give up some of their salary for more personal time. (Data extracted from “I'd Rather Give Up," USA Today, March 4, 2010. p. 1B.) Suppose you select a sample of 100 full-time female workers 22 to 35 years old. a. What is the probability that in the sample fewer than 50% would rather give up some of their salary for more per- sonal time? b. What is the probability that in the sample between 40% and 50% would rather give up some of their salary for more personal time? C. What is the probability that in the sample more than 40% would rather give up some of their salary for more per- sonal time? d. If a sample of 400 is taken, how does this change your answers to (a) through (c)?

Explanation / Answer

Solution:

Normal Distribution
Proportion p = 0.46
Standard Deviation s.d = sqrt(p*(1-p)/n) = sqrt(0.46*0.54/100) = 0.0498
Normal Distribution = Z= X- / sd ~ N(0,1)   
a)
P(X < 0.50) = (0.50-0.46)/0.0498
= 0.04/0.0498 = 0.8032
= P ( Z < 0.8032) From Standard Normal Table
= 0.7881
  
b)
To find P(a Z b) = F(b) - F(a)
P(X < 0.40) = (0.40-0.46)/0.0498
= -0.06/0.0498 = -1.2048
= P ( Z < -1.2048) From Standard Normal Table
= 0.1151
P(X < 0.5) = (0.50-0.46)/0.0498
= 0.04/0.0498 = 0.8032
= P ( Z < 0.8032) From Standard Normal Table
= 0.7881
P(0.40 < X < 0.50) = 0.7881- 0.1151 = 0.6730
  
c) P(X > 0.40) = (0.40-0.46)/0.0498
=-0.06/0.0498 = -1.2048
= P ( Z > -1.2048) From Standard Normal Table
= 0.8849

d) if n = 400 then standard error will be
standard error, SE = sqrt(p(1-p)/n)
= sqrt(0.46(1-0.46) / 400)
= 0.025
calculate parts a,b,c again using SE = 0.025 instead of 0.05

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