Question

A recent study showed that the modern working person experiences an average of 2.1 hours per day of distractions (phone calls, emails, impromptu visits, etc.). A random sample of 50 workers for a large corporation found that these workers were distracted an average of 1.8 hours per day and the population standard deviation was 20 minutes. Estimate the true mean population distraction time with 90% confidence, and compare your answer to that of the study.

Answer 1

Answer:

True mean population distraction time with 90% confidence is

C.I[103.34 ,112.66] at 108

Step-by-step explanation:

Given:

Study Average hrs =2.1 =2.1 *60=126 minutes

For sample mean =1.8 *60= 108 minutes

S.D=20  and n=50

To Find:

True mean population distraction time with 90% confidence,

Solution:

90% C.I. means 90 % will fall in true mean and other will not .

So for that calculating the

Standard error=S.D/Sqrt(n)

S.E=20/Sqrt(50)

S.E=2.828

For 90% Confidence interval Z=1.65

C.I= mean±Z*Standard error

C.I=108±1.65*2.828

C.I=108±4.662

Hence C.I will range from 103.34 to 112.66

Study mean =126 minutes .

Here  it ranges from 103.34 to 112.66

Answer 2

In this exercise we have to use the knowledge of probability to calculate this we will use percentage as:

$C.I= 108$

Given the following information in the text we find that:

• Study Average=126 minutes

• For sample mean =108 minutes

• S.D=20  and n=50

Then calculating the probability we find that:

$Standard error=S.D/\sqrt(n)\\ S.E=20/\sqrt(50)\\ S.E=2.828\\ Z=1.65\\ C.I=108+4.662$

See more about probability at brainly.com/question/795909