Solve Statistics Questions: North American TV Watching

[aikawa]

Homework Statement

Having problems solving some questions for Statistics class

The amount of time spent by North American adults watching TV per day is normally distributed with a mean of 6 hrs and a standard deviation of 1.5hrs.

a. What is the probability that a randomly selected North American adult watches TV for more than 7 hrs per day?

b. What is the probability that the average time watching TV by a random sample of 5 american adults is more than 7 hrs?

c. What is the probability that, in a random sample of 5 american adults, all watch TV for more than 7 hrs per day?

Homework Equations

P(X>7)

The Attempt at a Solution

Tried this but not sure if its correct

a. P(X>7)= (7-6/1.5)= .6667
=P(Z>.6667)= .2486
=0.5+.2486= .7486
The probablity that a randomly selected North American adult watches TV for more than 7 hrs per day is .7486%

Am i on the right path?

As for b and c I have no idea where to start.

Some advice/help would be much appreciated

 

[danago]

For (c), think about how you could use the binomial distribution to find the probability.

For (b), i think you could define a new random variable which is the average of the 5 observations i.e. Y = (X₁+X₂+...+X₅)/5, and then find P(Y>7), noting that the sum of normally distributed variables is also normally distributed.

 

[statdad]

Part 'b' does require you to use the distribution for $$\overline X$$.

In 'c', since you know the prob (from 'a') a single person watches for more than 7 hours per day, you're in good shape. Think about the multiplication rule now, since, for a random sample you have independence.

 

[Mark44]

For part a, you ended up with an incorrect answer. Here's some help in how you can get the right answer, and present it in an understandable way.

a. P(X>7)= (7-6/1.5)= .6667
=P(Z>.6667)= .2486
=0.5+.2486= .7486
The probablity that a randomly selected North American adult watches TV for more than 7 hrs per day is .7486%

Line 1: P(X > 7) = P(Z*sigma + mu > 7) = P(Z > (7 - mu)/sigma))
= P(Z > 2/3)

The relationship between your random variable X and the standard normal distrubution variable Z is Z = (X - mu)/sigma, so the relationship the other way is X = Z*sigma + mu.

Line 2: P(Z > 2/3) = 1 - P(Z < 2/3) = 1 - .7486 = .2514.
The table you used (I'm reasonably sure that you got your values from a table) gives the probability that Z is less than a particular value; IOW, that Z is between -infinity and the given value. From the table I used, the z-value is .67, which is not quite the same as 2/3. 2/3 is between .66 and .67, so for more precise results you can interpolate the area values for .66 and .67 to get a closer result.

Last line: You show the probability as .7486%, which is smaller than 1%. There are two errors here--the number itself, and showing the decimal probability as also a percentage. The work I show above gives the probability as about .25. As a percent, this would be 25%.
This says that the probability of a North American person watching more than 7 hours of TV a day is about 25%. Another way to say this is that about 25% of North American people watch more than 7 hours of TV a day.