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Based on all student records at Camford University, students spend an average of 5.5 hours per week playing organized sports. The population’s standard deviation is 2.2 hours per week. Based on a sample of 121 students, Healthy Lifestyles Incorporated (HLI) would like to apply the central limit theorem to make various estimates.
a) Compute the standard error of the sample mean.
b) What is the chance HLI will find a sample mean between 5 and 6 hours?
c) Calculate the probability that the sample mean will be between 5.3 and 5.7 hours.
d) How strange would it be to obtain a sample mean greater than 6.5 hours?

Respuesta :

Answer:

a)  [tex] SE=\sigma_{\bar X} = \frac{2.2}{\sqrt{121}}=0.2[/tex]

b) [tex] P(5 < \bar X <6)= P(\frac{5-5.5}{0.2}< Z<\frac{6-5.5}{0.2}) = P(-2.5< Z< 2.5)[/tex]

And we can find this probability like this:

[tex] P(-2.5< Z< 2.5)= P(Z<2.5)- P(Z<-2.5) = 0.994-0.006= 0.988[/tex]

c) [tex] P(5.3< \bar X <5.7)= P(\frac{5.3-5.5}{0.2}< Z<\frac{5.7-5.5}{0.2}) = P(-1< Z< 1)[/tex]

And we can find this probability like this:

[tex] P(-1< Z< 1)= P(Z<1)- P(Z<-1) = 0.841-0.159= 0.682[/tex]

d) [tex] P(\bar X >6.5)= 1-P(\bar X < 6.5) = 1-P(Z< \frac{6.5-5.5}{0.2}) = 1-P(Z< 5) = 1-0.999999 \approx 0[/tex]

So then since this probability is very near to 0 we can conclude that is a vert rare event obtain a sample mean greater than 6.5

Step-by-step explanation:

For this case let X the time spent per week playing organized sports and we knwo that:

[tex] E(X) = 5.5 , \sigma = 2.2[/tex]

And we select a sample of n=121 students

Part a

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".

From the central limit theorem we know that the distribution for the sample mean [tex]\bar X[/tex] is also normal and is given by:

[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})[/tex]

[tex]\bar X \sim N(\mu=5.5, \frac{2.2}{\sqrt{121}}=0.2)[/tex]

So then the standard error is [tex] SE=\sigma_{\bar X} = \frac{2.2}{\sqrt{121}}=0.2[/tex]

Part b

For this case we want to find this probability:

[tex] P(5<\bar X <6)[/tex]

And we can use the z score formula given by:

[tex] z= \frac{\bar X -\mu}{\sigma_{\bar X}}[/tex]

And using this formula we got:

[tex] P(5 < \bar X <6)= P(\frac{5-5.5}{0.2}< Z<\frac{6-5.5}{0.2}) = P(-2.5< Z< 2.5)[/tex]

And we can find this probability like this:

[tex] P(-2.5< Z< 2.5)= P(Z<2.5)- P(Z<-2.5) = 0.994-0.006= 0.988[/tex]

Part c

For this case we want to find this probability:

[tex] P(5.3< \bar X <5.7)[/tex]

And we can use the z score formula given by:

[tex] z= \frac{\bar X -\mu}{\sigma_{\bar X}}[/tex]

And using this formula we got:

[tex] P(5.3< \bar X <5.7)= P(\frac{5.3-5.5}{0.2}< Z<\frac{5.7-5.5}{0.2}) = P(-1< Z< 1)[/tex]

And we can find this probability like this:

[tex] P(-1< Z< 1)= P(Z<1)- P(Z<-1) = 0.841-0.159= 0.682[/tex]

Part d

For this case we can find [tex] P(\bar X >6.5)[/tex]

Using the z score and the complement rule we have this:

[tex] P(\bar X >6.5)= 1-P(\bar X < 6.5) = 1-P(Z< \frac{6.5-5.5}{0.2}) = 1-P(Z< 5) = 1-0.999999 \approx 0[/tex]

So then since this probability is very near to 0 we can conclude that is a vert rare event obtain a sample mean greater than 6.5

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