# Stats Questions

## 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?

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

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danago
Gold Member
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.

Homework Helper
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
Mentor
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.