# Math Help - Sampling distributions help

1. ## 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

2. Originally Posted by aikawa
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).

3. Originally Posted by aikawa
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)
$= (\text{result of A})^5$