Question

A survey from summer 2019 found that the average American adult spent 2.25 hours a day in front of a computer outside of work, with a standard deviation of 0.75 hours. The survey also found that the population data is most likely normally distributed. a) What portion of adults are in front of a computer for less than 0 hours a day, outside of work? b) Medical professions recommend spending no more than 2 hours in front of a computer each day. What portion of the population is on their computer more than the recommended amount? c)What portion of adults spend between 1 and 5 hours a day in front of a compute reach day, outside of work?

Answer 1

a)0.135%

b) 36.944%

c) 95.209%

Explanation:

We solve the above question using z score formula

z = (x-μ)/σ, where

x is the raw score

μ is the population mean = 2.25 hours

σ is the population standard deviation = 0.75 hours

a) What portion of adults are in front of a computer for less than 0 hours a day, outside of work?

x < 0 hours

Hence,

z = 0 - 2.25/0.75

= -3

Probability value from Z-Table:

P(x<0) = 0.0013499

Converting to Percentage

= 0.0013499 × 100

= 0.13499%

= 0.135%

b) Medical professions recommend spending no more than 2 hours in front of a computer each day. What portion of the population is on their computer more than the recommended amount?

No more than 2 hours

x ≤ 2 hours

x < 2 hours

Hence,

z = 2 - 2.25/0.75

= -0.33333

Probability value from Z-Table:

P(x≤ 2) = 0.36944

Converting to Percentage

= 0.36944 × 100

= 36.944%

c)What portion of adults spend between 1 and 5 hours a day in front of a compute reach day, outside of work?

For x = 1 hour

z = 1 - 2.25/0.75

= -1.66667

P-value from Z-Table:

P(x= 1) = 0.04779

z = 5 - 2.25/0.75

= 3.66667

P-value from Z-Table:

P(x = 5) = 0.99988

Hence,

0.99988 - 0.04779

= 0.95209

Converting to Percentage

= 0.95209 × 100

= 95.209%