Results 1 to 3 of 3

Math Help - Sampling distributions help

  1. #1
    Newbie
    Joined
    May 2009
    Posts
    2

    Sampling distributions help

    Hello need some help on few questions

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

    A. what is the probability that a random selected North American adult watches tv for more than 7 hrs?

    B. What is the probability that the average time watching tv by a random sample of five adults is more than 7 hrs?

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

    Here is what i came up with

    A. Z= 7-6/1.5= .6667
    P(Z>.6667)=.2486
    =0.5 + .2486= .7486

    B. 7-6/.67= 1.49
    P(Z>1.49)= 0.5- .4319= .0681

    C. Cannot come up with a solution

    Am i on the right track? doing the calculation correct? When Z>X is it always going to me subtracting and when Z<X addition?

    Some help/advice is much appreciated
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by aikawa View Post
    Hello need some help on few questions

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

    A. what is the probability that a random selected North American adult watches tv for more than 7 hrs?

    B. What is the probability that the average time watching tv by a random sample of five adults is more than 7 hrs?

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

    Here is what i came up with

    A. Z= 7-6/1.5= .6667
    P(Z>.6667)=.2486
    =0.5 + .2486= .7486 Mr F says: This is clearly wrong. Since 7 is greater than the mean, how can Pr(X > 7) be greater than 0.5 ....? I get 0.2525.

    B. 7-6/.67= 1.49
    P(Z>1.49)= 0.5- .4319= .0681 Mr F says: Correct.

    C. Cannot come up with a solution

    Am i on the right track? doing the calculation correct? When Z>X is it always going to me subtracting and when Z<X addition?

    Some help/advice is much appreciated
    C. Let Y be the random variable number of adults in the sample that watch TV for more than 7 hours.

    Y ~ Binomial(n = 5, p = answer to A.)

    Calculate Pr(Y = 5).
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    May 2009
    Posts
    39
    Quote Originally Posted by aikawa View Post
    Hello need some help on few questions

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

    A. what is the probability that a random selected North American adult watches tv for more than 7 hrs?

    B. What is the probability that the average time watching tv by a random sample of five adults is more than 7 hrs?

    C. What is the probability that in a random sample of five adults, all watch TV for more than 7 hrs per day?
    Denote X is amount of time that a North American uses for watching tv. I have X~N(6; 1.5^2)
    A. We calculate  P\{X > 7\} = 1 - P\{X<7\}=1-P\{\frac{X-6}{1.5} <\frac{7-6}{1.5}\}
     =1 - P\{U < 0.66\}=1-0.7486=0.2514
    B.  \bar{X} = \frac{X_1 +X_2+X_3+X_4+X_5}{5}
     \bar{X} -N(6; 1.5^2)
    We need calculate  P\{\bar{X} > 7\}
    we obtain the same result as A
    C. the request is equivalent
     P\{ X_1 > 7 ; X_2 >7; X_3>7;X_4>7;X_5>7\}
     = P\{ X_1 > 7\} P\{ X_2 > 7\} P\{ X_3 > 7\}P\{ X_4 > 7\}P\{ X_5 > 7\} (because of independent)
     = (\text{result of A})^5
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Sampling Distributions
    Posted in the Statistics Forum
    Replies: 2
    Last Post: June 15th 2014, 05:49 PM
  2. Sampling distributions
    Posted in the Advanced Statistics Forum
    Replies: 2
    Last Post: September 2nd 2010, 10:15 PM
  3. Sampling Distributions Help
    Posted in the Advanced Statistics Forum
    Replies: 2
    Last Post: February 2nd 2010, 06:12 PM
  4. Sampling Distributions
    Posted in the Statistics Forum
    Replies: 0
    Last Post: January 31st 2010, 01:20 AM
  5. Sampling distributions
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: October 22nd 2007, 10:28 AM

Search Tags


/mathhelpforum @mathhelpforum