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 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 100 100 working women. Answer parts​ (a) through​ (d). a. What is the probability that in the sample fewer than 45 45​% say that having a flexible work schedule is either very important or extremely important to their career​ success? nothing ​(Round to four decimal places as​ needed.) b. What is the probability that in the sample between 35 35​% and 45 45​% say that having a flexible work schedule is either very important or extremely important to their career​ success? nothing ​(Round to four decimal places as​ needed.) c. What is the probability that in the sample more than 40 40​% say that having a flexible work schedule is either very important or extremely important to their career​ success? nothing ​(Round to four decimal places as​ needed.) d. If a sample of 400 400 is​ taken, how does this change your answers to​ (a) through​ (c)? The probability that in the sample fewer than 45 45​% say that having a flexible work schedule is either very important or extremely important to their career success is nothing . The probability that in the sample between 35 35​% and 45 45​% say that having a flexible work schedule is either very important or extremely important to their career success is nothing . The probability that in the sample more than 40 40​% say that having a flexible work schedule is either very important or extremely important to their career success is nothing .

Answer 1

p = 0.42

n = 100

= p = 0.42

= sqrt(p(1 - p)/n)

      = sqrt(0.42 * (1 - 0.42)/100)

      = 0.0494

a) P( < 0.45)

= P(( - )/ < (0.45 - )/)

= P(Z < (0.45 - 0.42)/0.0494)

= P(Z < 0.61)

= 0.7291

b) P(0.35 < < 0.45)

= P((0.35 - )/ < ( - )/ < (0.45 - )/)

= P((0.35 - 0.42)/0.0494 < Z < (0.45 - 0.42)/0.0494)

= P(-1.42 < Z < 0.61)

= P(Z < 0.61) - P(Z < -1.42)

= 0.7291 - 0.0778

= 0.6513

c) P( > 0.4)

= P(( - )/ > (0.4 - )/)

= P(Z > (0.4 - 0.42)/0.0494)

= P(Z > -0.40)

= 1 - P(Z < -0.40)

= 1 - 0.3446

= 0.6554

d) n = 400

p = 0.42

= p = 0.42

= sqrt(p(1 - p)/n)

      = sqrt(0.42 * (1 - 0.42)/400)

      = 0.0248

i) P( < 0.45)

= P(( - )/ < (0.45 - )/)

= P(Z < (0.45 - 0.42)/0.0248)

= P(Z < 1.21)

= 0.8869

ii) P(0.35 < < 0.45)

= P((0.35 - )/ < ( - )/ < (0.45 - )/)

= P((0.35 - 0.42)/0.0248 < Z < (0.45 - 0.42)/0.0248)

= P(-2.82 < Z < 1.21)

= P(Z < 1.21) - P(Z < -2.82)

= 0.8869 - 0.0024

= 0.8845

iii) P( > 0.4)

= P(( - )/ > (0.4 - )/)

= P(Z > (0.4 - 0.42)/0.0248)

= P(Z > -0.81)

= 1 - P(Z < -0.81)

= 1 - 0.2090

= 0.7910