Question

The distribution of the length of waiting time makes the passport approach the normal distribution with an average of 280 seconds and the standard deviation of 90 seconds. It is known that 75% of all passport owners make it online. A random sample of 50 people was taken.

Determine the chance of the proportion of people making passports online between 55% and 65%.

Answer 1

P = 0.75

n = 50

= 0.75

= sqrt(p(1 - p)/n)

= sqrt(0.75 * (1 - 0.75)/50)

= 0.0612

P(0.55 < < 0.65)

= P((0.55 - )/ < ( - )/ < (0.65 - )/)

= P((0.55 - 0.75)/0.0612 < Z < (0.65 - 0.75)/0.0612)

= P(-3.27 < Z < -1.63)

= P(Z < -1.63) - P(Z < -3.27)

= 0.0516 - 0.0005

= 0.0511