Do ork- Michael Beale Google Chrome Secure https://www.mathud.com/Stadent/Player
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Do ork- Michael Beale Google Chrome Secure https://www.mathud.com/Stadent/Player DS 2333 Fall 2017 Michael Beale 11/11/17 7:50 Homework: Chapter 7: Sec. 1, 2,&3 Score: 0 of 1 pt 50r 5 (5 complete) HW Score: 41 84%, 293 of 7.3.14-T global sesearch study found that the majority of today's working women would prefr a bettor work ate balance to an increased salary Oie of the most Questor, Heip contrbutors to work-le balance identrned sextilly-aff, 45% nt to ther caroer success Suppose you select a sample of 100 working women Answer parts (a) through (d) awh tstre protataty that nthe sample fower t succoss? han 52% say that having dfexble work schende is ether very important or extamaly "mportant to her canoer (Round to tour decimal places as needed) Enter your answer in the answer box and then click Check Answer 3 aring- Clear AllExplanation / Answer
7.3.14.
the PDF of normal distribution is = 1/ * 2 * e ^ -(x-u)^2/ 2^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd/sqrt(n) ~ N(0,1)
proportion ( p ) = 0.45
standard Deviation ( sd )= sqrt(PQ/n) = sqrt(0.45*0.55/100)
=0.0497
a.
P(X < 0.52) = (0.52-0.45)/0.0497
= 0.07/0.0497= 1.4085
= P ( Z <1.4085) From Standard Normal Table
= 0.9205
b . the probability that in the sample between 41% and 52% say that having a flexible work schedule is either very important or extremely important to their career success
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 0.41) = (0.41-0.45)/0.0497
= -0.04/0.0497 = -0.8048
= P ( Z <-0.8048) From Standard Normal Table
= 0.21046
P(X < 0.52) = (0.52-0.45)/0.0497
= 0.07/0.0497 = 1.4085
= P ( Z <1.4085) From Standard Normal Table
= 0.9205
P(0.41 < X < 0.52) = 0.9205-0.21046 = 0.71
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