Question

The amount of time travellers at an airport spend with customs officers has a mean of  μ =33 μ =33 seconds and a standard deviation of  σ =15 σ =15 seconds. For a random sample of 35 travellers, what is the probability that their mean time spent with customs officers will be:

Standard Normal Distribution Table

a. Over 30 seconds?

Round to four decimal places if necessary

b. Under 35 seconds?

Round to four decimal places if necessary

c. Under 30 seconds or over 35 seconds?

Round to four decimal places if necessary

Answer 1

Solution :

Given that,

mean = = 33

standard deviation = = 15

n = 35

= 33

= / n = 15 35 = 2.5355

a ) P (> 30 )

= 1 - P ( < 30 )

= 1 - P ( - / ) < ( 30 - 33 / 2.5355)

= 1 - P ( z <- 3 / 2.5355 )

= 1 - P ( z < -1.18 )

Using z table

= 1 - 0.1190

= 0.8810

Probability = 0.8810

b ) P( < 35 )

P ( - / ) < ( 35 - 33 / 2.5355)

P ( z < 2 / 2.5355 )

P ( z < 0.79)

= 0.7852

Probability = 0.7852

c ) P( < 30 )

P ( - / ) < ( 30 - 33 / 2.5355)

P ( z < -3 / 2.5355 )

P ( z < -1.18)

= 0.1190

P ( > 35 )

= 1 - P ( < 35 )

= 1 - P ( - / ) < ( 35 - 33 / 2.5355)

= 1 - P ( z < 2 / 2.5355 )

= 1 - P ( z < 0.79 )

Using z table

= 1 - 0.7852

= 0.2148

Probability = 0.1190 + 0.2148 = 0.3338