Question

Bonnectmath.com/alekscgl/x/sl.exe/1o_u-IgNsikr7;8P3JH-IV-TghIJBUeBr8WON7n|F20J6p6bnJC_JULy3RSLYwaekZrWorMFDKT7SWNMOGO Stress at work: In a poll conducted by the General Social Survey, 80% of respondents said that their jobs were sometimes or always stressful. Two hundred workers are chosen at random. Use the TI-84 Plus calculator as needed. Round least four decimal places. (a) Approximate the probability that 165 or fewer workers find their jobs stressful. (b) Approximate the probability that more than 150 workers find their jobs stressful. (c) Approximate the probability that the number of workers who find their jobs stressful is between 153 and 160 inclusive. Part 1 of 3 The probability that 165 or fewer workers find their jobs stressful is 8340 Part 2 of 3 The probability that more than 150 workers find their jobs stressful is .9686 Part 3 of 3 The probability that the number of workers who find their jobs stressful is between 153 and 160 inclusive is .4441

Answer 1

Solution :

Given that n = 200 , p = 0.80

=> q = 1 - p = 0.20

=> mean μ = n*p

= 200*0.80

= 160

=> standard deviation σ = sqrt(n*p*q)

= sqrt(200*0.80*0.20)

= 5.6569

(a)
=> P(x <= 165) = P((x - μ)/σ <= (165 - 160)/5.6569)

= P(Z <= 0.8839)

= 0.8106

(b)
=> P(x > 150) = P((x - μ)/σ > (150 - 160)/5.6569)

= P(Z > -1.7678)

= 0.9616

(c)
=> P(153 <= x <= 160) = P((153 - 160)/5.6569 < (x - μ)/σ < (160 - 160)/5.6569)

= P(-1.2374 < Z < 0)

= 0.3925