Question

The mean of hours that the average person watches television each day is 4.18 hours with a standard deviation of 1.19 hours. Find probability that someone watches between 3 and 5 hours a day

Answer 1

Answer:

$z = \frac{3-4.18}{1.19}=-0.992$

$z = \frac{5-4.18}{1.19}=0.689$

And we can find this probability with this difference:

$P(-0.992$

And then we can conclude that the probability that someone watches between 3 and 5 hours a day is approximately 0.591 using a normal distribution

Explanation:

For this case we can define the random variable X as "hours that a person watches television". For this case we don't have the distribution for X but we have the following parameters:

$\mu = 4.18,\sigma =1.19$

We can assume that the distribution for X is normal

$X \sim N(\mu = 4.18 , \sigma =1.19)$

And we want to find this probability:

$P(3$

And we can use the z score formula given by:

$z=\frac[X- \mu}{\sigma}$

And we can find the z score for each limit and we got:

$z = \frac{3-4.18}{1.19}=-0.992$

$z = \frac{5-4.18}{1.19}=0.689$

And we can find this probability with this difference:

$P(-0.992$

And then we can conclude that the probability that someone watches between 3 and 5 hours a day is approximately 0.591 using a normal distribution