Could someone help me with this question?

Question

The number of hours per week adults in the U.S. spend on home computers is normally distributed, with a mean of 6 hours and a standard deviation of 2 hour. An adult in the U.S. selected.

1.  Find the probability that the hours spent on the home computer by the adult are less than 2.5 hours per week.

1.  Find the probability that the hours spent on the home computer by the adult are between 2.5 and 7.5 hours per week.

1.  Find the probability that the hours spent on the home computer by the adult are more than 7.5 hours per week.

1. What percent of the adults spend more than 2 hours per week on a home computer?

1. If 350 adults in the U.S. were randomly selected, about how many would you expect to say they spend less than 3 hours per week on a home computer?

Answer 1

This problem boils down to z scores which are calculated as z = (x-µ)/σ  where µ = mean and σ is standard deviation. After this we need to look at the lookup table to find the area under the normal distribution to the left of the z score (in other words the probability that x < c). Here is a z-score look up table and a good explanation: http://www.z-table.com

1) z = 2.5-6/2 z = -1.75

p=.0401

2)z₁ = -0.5 z₂ = 7.5-6/2 = 0.75

p = P(z₂)-P(z₁)

p = 0.7734 - .0401 = 0.7333

3) z = 0.75

p(z>0.75) = 1 - 0.7734 = 0.2266

4) z = 2-6/2 = -2

P(z > .0228) = 1 - .0228 = 0.9772

5) z = 3-6/2 = -1

p(z=-1) = .1587

.1587*350 = 55.545 = 55