Question

A global research study found that the majority of​ today's working women would prefer a better​ work-life balance to an increased salary. One of the most important contributors to​ work-life balance identified by the survey was​ "flexibility," with 42​% of women saying that having a flexible work schedule is either very important or extremely important to their career success. Suppose you select a sample of 100 working women. Answer parts​ (a) through​ (d).

a. What is the probability that in the sample fewer than 45​% say that having a flexible work schedule is either very important or extremely important to their career​ success?

ANSWER= .7283

​(Round to four decimal places as​ needed.)

b. What is the probability that in the sample between 34​% and 45​% say that having a flexible work schedule is either very important or extremely important to their career​ success?

​(Round to four decimal places as​ needed.)

Answer 1

Solution:

Given that p = 0.42, q = 1 - p = 0.58, n = 100

Here , np = 0.42*100 = 42 and nq = 0.58*100 = 58

By using normal approximation
a) P(p̂ < 0.45) = P[p̂ -p/sqrt(p(1-p)/n) < 0.45 - 0.42/sqrt(0.42(1-0.42)/100]
= P[Z < 0.45 - 0.42/sqrt(0.42(1-0.42)/100]
= P[Z < 0.03/0.0493]
= P[Z < 0.6078]
= 0.7291 (From the normal distribution table)
------------------------------------------------------------------------------------
b) P(0.34 < p̂ < 0.45) = P[0.34 - 0.42/sqrt(0.42(1-0.42)/100 < p̂ -p/sqrt(p(1-p)/n) < 0.45 - 0.42/sqrt(0.42(1-0.42)/100]
= P[0.34 - 0.42/sqrt(0.42(1-0.42)/100 < Z < 0.45 - 0.42/sqrt(0.42(1-0.42)/100]
= P[-0.08/0.0493 < Z < 0.03/0.0493]
= P[-1.6209 < Z < 0.6078]
= P(Z < 0.6078 ) - P(Z< -1.6209 )
= 0.7291 - (1-0.9474)
= 0.7291 - 0.0526
= 0.6765