Question

1. According to Statistics Canada, about 71% of commuters in Canada drive to work. Suppose 150 Canadian commuters are randomly sampled.

(a) What is the probability that fewer than 105 commuters drive to work?

(b) What is the probability that between 110 and 120 (inclusive) commuters drive to work?

(c) What is the probability that more than 95 commuters drive to work?

Could you please show all steps, thank you.

Answer 1

Answer)

P = 0.71

N = 150

First we need to check the conditions of normality that is if n*p and n*(1-p) both are greater than 5 or not

N*p = 106.5

N*(1-p) = 43.5

Both the conditions are met so we can use standard normal z table to estimate the probability

Z = (x - mean)/s.d

Mean = n*p = 106.5

S.d = √{n*p*(1-p)} = 5.55742746241

A)

P(x<105)

By continuity correction

P(x<104.5)

Z = (104.5 - 106.5)/5.55742746241 = -0.36

From z table, P(z<-0.36) = 0.3594

B)

P(110<x<120) = P(x<120.5) - P(x<109.5) by continuity correction

P(x<120.5)

Z = 2.52

From z table, P(z<2.52) = 0.9941

P(x<109.5)

Z = 0.54

From z table, P(z<0.54) = 0.7054

Required probability is 0.9941 - 0.7054 = 0.2887

C)

P(x>95)

By continuity correction

P(x>95.5)

Z = -1.98

From z table, P(z>-1.98) =

0.9761