Question

Question 5 The speed at which you can log into a website through a smartphone is an important quality characteristic of that website. In a recent test, the mean time to log into the hertz website through a smartphone was 7.524 seconds. (Data extracted from N. Trejos, ―Travellers Have No Patience for Slow Mobile Sites, ―USA Today, April 4, 2012 , p.3B.) Suppose that the download time is normally distributed, with a standard deviation of 1.7 seconds. What is the probability that a download time is: a. Less than 2 seconds? b. Between 1.5 and 2.5 seconds? c. Above 1.8 seconds? d. 99% of the download times are slower (higher) than how many seconds? e. 95% of the download times are between what two values, symmetrically distributed around the mean?

Answer 1

Solution:

Given:

Mean = seconds

Standard deviation = seconds

Part a) the probability that a download time is Less than 2 seconds.

P( X < 2 ) =..........?

Find z score:

Thus we get:

P( X < 2 ) = P( Z < -3.25)

Look in z table for z = -3.2 and 0.05 and find area.

P( Z < -3.25) = 0.0006

Thus

P( X < 2 ) = P( Z < -3.25)

P( X < 2 ) = 0.0006

Part b)   the probability that a download time is Between 1.5 and 2.5 seconds.

P( 1.5 < X < 2.5 ) =............?

z score for x = 1.5 and for x = 2.5

Thus we get:

P( 1.5 < X < 2.5 ) = P( -3.54 < Z < -2.96 )

P( 1.5 < X < 2.5 ) = P( Z < -2.96 ) - P( Z < -3.54)

Look in z table for z = -2.9 and 0.06 as well as for z = -3.5 and 0.04 and find area:

P( Z < -2.96) = 0.00154

P( Z < -3.54) = 0.00020

Thus

P( 1.5 < X < 2.5 ) = P( Z < -2.96 ) - P( Z < -3.54)

P( 1.5 < X < 2.5 ) = 0.00154 - 0.00020

P( 1.5 < X < 2.5 ) = 0.00134

P( 1.5 < X < 2.5 ) = 0.0013

Part c) the probability that a download time is above 1.8 seconds

P( X > 1.8) =...........?

z score for x = 1.8

P( X > 1.8) = P( Z > -3.37)

P( X > 1.8) =1 - P( Z < -3.37)

P( Z < -3.37) = 0.0004

Thus

P( X > 1.8) =1 - P( Z < -3.37)

P( X > 1.8) =1 - 0.0004

P( X > 1.8) = 0.9996

Part d) 99% of the download times are slower (higher) than how many seconds?

P( X > x ) = 0.99

Find z such that: P( Z > z) = 0.99

that is find z such that:

P( Z < z ) = 1 - 0.99

P( Z < z ) = 0.01

Look in z table for area = 0.01 and find z.

Area 0.0099 is closest to 0.0100 and it corresponds to -2.33

Thus we use following formula:

thus 99% of the download times are slower (higher) than 3.563 seconds.

Part e) 95% of the download times are between what two values, symmetrically distributed around the mean?

That is we have to find two x values such that:

P(x1 < X < x2) =0.95

0.95 area is middle area

let c = middle area = 0.95 then

Find Area = ( 1 + c ) / 2 = ( 1 + 0.95) /2 = 1.95 / 2 = 0.9750

Look in z table for Area = 0.9750 or its closest area and find z value.

Area = 0.9750 corresponds to 1.9 and 0.06 , thus z critical value = 1.96

That is : Z = 1.96

thus we have two z values : -1.96 and 1.96

Now use following formula:

and

Thus 95% of the download times are between 4.192 and 10.856 seconds.