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Homework Help: Stats Questions

  1. May 25, 2009 #1
    1. The problem statement, all variables and given/known data
    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?

    2. Relevant equations


    3. 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
  2. jcsd
  3. May 25, 2009 #2


    User Avatar
    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.
  4. May 25, 2009 #3


    User Avatar
    Homework Helper

    Part 'b' does require you to use the distribution for [tex] \overline X [/tex].

    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.
  5. May 26, 2009 #4


    Staff: 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.
    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.
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