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?

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?

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)
$= (\text{result of A})^5$