Question

Technology advancements have become a driving force for national and economic growth to transact across the global. Service providers have become so alarmed with the rate at which customers are able to download a web page. In one of the optic fiber survey, it was noted that the most customers tend to experience a mean download time of a resource web page is normally distributed to 8.5 seconds. After analysis by the census office, the data obtained had a standard deviation of 4.5 seconds.

1. What probability of page downloads take less than 5 seconds?
2. What is the probability that the download time will be between 5 and 11 seconds?
3. How many seconds elapse before 25% of downloads are complete?
4. If 10 downloads are randomly selected, how many downloads are expected to take more than 12 seconds
5. What is the probability that the download time will be between 10 and 12 seconds

Answer 1

Solution :

Given that ,

mean = = 8.5

standard deviation = = 4.5

P(x < 5 ) = P[(x - ) / < ( 5 - 8.5) / 4.5 ]

= P(z < -0.78 )

Using z table,

= 0.2177

Probability = 0.2177

( b )

P( 5 < x < 11 )

= P[( 5 - 8.5 ) / 4.5 ) < (x - ) /  < ( 11 - 8.5) / 4.5) ]

= P( -0.78 < z < 0.56 )

= P(z < 0.56 ) - P(z < -0.78 )

Using z table,

= 0.7123 - 0.2177

= 0.4946

Probability = 0.4946

( c )

The z distribution of the  25% is ,

P(Z < z) = 25%

= P(Z < z ) = 0.25

= P(Z < -0.674 ) = 0.25

z = -0.674

Using z-score formula,

x = z * +

x = -0.674 * 4.5 + 8.5

x = 5.467

Answer = x = 5

( d )

P(x > 12 ) = 1 - P( x < 12 )

=1- P[(x - ) / < ( 12 - 8.5) / 4.5 ]

=1- P(z < 0.78 )

Using z table,

= 1 - 0.7823

= 0.2177

= 0.2177 * 10 = 2.177

= 2.177

Answer = 2

( e )

P( 10 < x < 12 )

= P[( 10 - 5.8 ) / 4.5 ) < (x - ) /  < ( 12 - 8.5 ) / 4.5 ) ]

= P( 0.33 < z < 0.78)

= P(z < 0.78 ) - P(z < 0.33 )

Using z table,

= 0.7823 - 0.6293

= 0.1530

Probability = 0.1530