Question

Just Part B please

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 45​% 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 51​% say that having a flexible work schedule is either very important or extremely important to their career​ success? 0.8869 ​(Round to four decimal places as​ needed.)

b. What is the probability that in the sample between 41​% and 51​% 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.)

Answer 1

Probability of women saying having a flexible work schedule is either very important or extremely important to their career success, P = 45% = 0.45

sample size, n = 100

the variance of women said having a flexible work schedule is either very important or extremely important to their career success = p*(1-p)/n

variance = 0.45*(1-0.45)/100

variance = 0.45*0.55/100

variance = 0.2475/100

variance = 0.002475

standard deviation = sqrt(variance)

standard deviation = sqrt(0.002475)

standard deviation = 0.0497

We need to find probability of women say having a flexible work schedule is either very important or extremely important to their career success between 41% and 51%

P[ 0.41 < p < 0.51 ] = P[ ( 0.41 - P )/sd < ( p - P )/sd < ( 0.51 - P )/sd ]

P[ 0.41 < p < 0.51 ] = P[ ( 0.41 - 0.45 )/0.0497 < ( p - 0.45 )/0.0497 < ( 0.51 - 0.45 )/0.0497 ]

P[ 0.41 < p < 0.51 ] = P[ -0.80 < Z < 1.21 ]

P[ 0.41 < p < 0.51 ] = P[ Z < 1.21 ] - P[ Z < -0.8 ]

P[ 0.41 < p < 0.51 ] = 0.8869 - 0.2119

P[ 0.41 < p < 0.51 ] = 0.675