Respuesta :
Answer:
[tex] z = \frac{3-4.18}{1.19}=-0.992[/tex]
[tex] z = \frac{5-4.18}{1.19}=0.689[/tex]
And we can find this probability with this difference:
[tex] P(-0.992<z<0.689) = P(z<0.689) -P(z<-0.992) =0.752 -0.161=0.591[/tex]
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:
[tex]\mu = 4.18,\sigma =1.19[/tex]
We can assume that the distribution for X is normal
[tex] X \sim N(\mu = 4.18 , \sigma =1.19)[/tex]
And we want to find this probability:
[tex] P(3 <X<5)[/tex]
And we can use the z score formula given by:
[tex] z=\frac[X- \mu}{\sigma}[/tex]
And we can find the z score for each limit and we got:
[tex] z = \frac{3-4.18}{1.19}=-0.992[/tex]
[tex] z = \frac{5-4.18}{1.19}=0.689[/tex]
And we can find this probability with this difference:
[tex] P(-0.992<z<0.689) = P(z<0.689) -P(z<-0.992) =0.752 -0.161=0.591[/tex]
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
