Question

4. A study of drivers reveals that, when lost, 45% will stop and ask for directions, 30% will consult a map, and 25% will continue driving until the location has been determined. Suppose that a sample of 200 drivers was asked to report what they do when lost. Find the following probabilities.

a. At least 100 stop and ask directions

b. at most 55 continue driving

c. between 50 and 75 (inclusive) consult a map

Answer 1

Answer)

As the sample size is large enough

We can use normal approximation here

Z = (x - mean)/s.d

Mean = n*p

S.d = √{n*p*(1-p)}

A)

P = 0.45

N = 200

Mean = 200*0.45 = 90

S.d = √{200*0.45*0.55} = 7.03562363973

P(>=100)

P(x>99.5) by continuity correction

Z = (99.5 - 90)/7.03562363973 = 1.35

From z table, P(z>1.35) = 0.0885

B)

P(<=55)

P(x<55.5)

Mean = n*p = 200*0.25 = 50

S.d = √{200*0.25*0.75} = 6.12372435695

Z = (55.5 - 50)/6.12372435695 = 0.9

From z table, P(z<0.9) = 0.8159

C)

P(50<x<75)

By.continuity correction

P(49.5<x<75.5) = p(x<75.5) - p(x<49.5)

Mean = 200*0.3 = 60

S.d = √{200*0.3*0.7} = 6.48074069840

P(x<75.5)

Z = 2.39

From z table, P(z<2.39) = 0.9916

P(x<49.5)

Z = (49.5 - 60)/6.48074069840 = -1.62

From z table, P(z<-1.62) = 0.0526

Required probability is 0.9916 - 0.0526 = 0.939