Respuesta :
Answer:
a) [tex] P(\bar X <10)[/tex]
We can use the z score given by:
[tex] z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
And if we find the z score we got:
[tex] z = \frac{10-8.1}{\frac{1.5}{\sqrt{49}}} = 8.867[/tex]
And we want this probability:
[tex] P(\bar X <10) =P(Z<8.867) \approx 1[/tex]
b) [tex] z = \frac{5-8.1}{\frac{1.5}{\sqrt{49}}} = -14.467[/tex]
[tex] z = \frac{10-8.1}{\frac{1.5}{\sqrt{49}}} = 8.867[/tex]
And we want this probability:
[tex] P(5< \bar X <10) =P(Z<8.867)-P(Z<-14.467) \approx 1[/tex]
c) [tex] z = \frac{6-8.1}{\frac{1.5}{\sqrt{49}}} = -9.8[/tex]
And we want this probability:
[tex] P(\bar X <6) =P(Z<-9.8) \approx 0[/tex]
Step-by-step explanation:
Assuming a sample size of n =49
For this case we define the random variable X ="amount of time that a customer spends waiting at an airport check-in counter " and we know:
[tex] E(X) = 8.1 , \sigma = 1.5[/tex]
Previous concepts
The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
Solution to the problem
From the central limit theorem we know that the distribution for the sample mean [tex]\bar X[/tex] is given by:
[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})[/tex]
Part a
We want this probability:
[tex] P(\bar X <10)[/tex]
We can use the z score given by:
[tex] z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
And if we find the z score we got:
[tex] z = \frac{10-8.1}{\frac{1.5}{\sqrt{49}}} = 8.867[/tex]
And we want this probability:
[tex] P(\bar X <10) =P(Z<8.867) \approx 1[/tex]
Part b
We want this probability:
[tex] P(5< \bar X <10)[/tex]
We can use the z score given by:
[tex] z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
And if we find the z score we got:
[tex] z = \frac{5-8.1}{\frac{1.5}{\sqrt{49}}} = -14.467[/tex]
[tex] z = \frac{10-8.1}{\frac{1.5}{\sqrt{49}}} = 8.867[/tex]
And we want this probability:
[tex] P(5< \bar X <10) =P(Z<8.867)-P(Z<-14.467) \approx 1[/tex]
Part c
We want this probability:
[tex] P(\bar X <6)[/tex]
We can use the z score given by:
[tex] z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
And if we find the z score we got:
[tex] z = \frac{6-8.1}{\frac{1.5}{\sqrt{49}}} = -9.8[/tex]
And we want this probability:
[tex] P(\bar X <6) =P(Z<-9.8) \approx 0[/tex]
