Question

According to the U.S. Bureau of the Census, about 75% of commuters in the United States drive to work alone. Suppose 150 U.S. commuters are randomly sampled. (a) What is the probability that fewer than 105 commuters drive to work alone? (b) What is the probability that between 112 and 124 (inclusive) commuters drive to work alone? (c) What is the probability that more than 95 commuters drive to work alone? (Round the values of σ to 4 decimal places. Round the values of z to 2 decimal places. Round your answers to 4 decimal places.)

Answer 1

solution:-

given that n = 150 , p = 0.75

mean (np) = 0.75*150 = 112.5

standard deviation = sqrt(n*p*q) = sqrt(150*0.75*(1-0.75))

= 5.3033

(a) P(X < 105) = P(Z < (105-112.5)/5.3033)

= 1-P(Z < 1.41)

= 1 - 0.9207

= 0.0793

(b) P(112 < X < 124) = P((112-112.5)/5.3033 < Z < (124-112.5)/5.3033)

= P(-0.09 < Z < 2.17)

= P(Z< 2.17) - P(Z < -0.09)

= 0.9850 - 0.4641

= 0.5209

(c) P(X > 95) = P(Z > (95-112.5)/5.3033)

= P(Z > -3.30)

= P(Z < 3.30)

= 0.9995