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Expert-verifiedLength of job tenure. Researchers at the Terry College ofBusiness at the University of Georgia sampled 344 business students and asked them this question: “Over the course of your lifetime, what is the maximum number of years you expect to work for any one employer?” The sample resulted in x = 19.1 years. Assume that the sample of students was randomly selected from the 6,000 undergraduate students atthe Terry College and that = 6 years.
In Terry College of Business at the University of Georgia, some researchers took a sample of 344 students from a population of 6000 undergraduate students randomly. Then they questioned the sampled students that what is the number of maximum years they expect to work for any one employer.
The survey resulted that x=19.1 years. The standard deviation $\sigma =6\text{\hspace{0.33em}}years$.
The sampling distribution of $\overline{x}$ is described as,
Mean ${\mu}_{\overline{x}}=\mu $
Standard deviation ${\sigma}_{\overline{x}}=\frac{\sigma}{\sqrt{n}}=\frac{6}{\sqrt{344}}\approx 0.323$
z-score $z=\frac{\overline{x}-{\mu}_{\overline{x}}}{{\sigma}_{\overline{x}}}=\frac{\overline{x}-\mu}{\raisebox{1ex}{$\sigma $}\!\left/ \!\raisebox{-1ex}{$\sqrt{344}$}\right.}$
We have the mean of the 6000 undergraduate students, that is $\mu =18.5$.
Here, $n=344$. So, n is greater than 30. We can say that it is a relatively large sample size.
Therefore, by the Central Limit Theorem we can say that $\overline{X}~N\left(18.5,0.323\right)$.
Thus,
$\begin{array}{c}P\left(\overline{X}>19.1\right)=normalcdf\left(19.1,{9}^{99},18.5,0.323\right)\\ \approx 0.0318\end{array}$
We have the mean of the 6000 undergraduate students, that is $\mu =19.5$.
Here, $n=344$. So, n is greater than 30. We can say that it is a relatively large sample size.
Therefore, by the Central Limit Theorem we can say that $\overline{X}~N\left(19.5,0.323\right)$.
Thus,
$\begin{array}{c}\mathrm{P}\left(\overline{\mathrm{X}}>19.1\right)=\mathrm{normalcdf}\left(19.1,{9}^{99},19.5,0.323\right)\\ \approx 0.8919\end{array}$
There is given that $\mathrm{P}\left(\overline{\mathrm{X}}>19.1\right)=0.5$. So, we can conclude that 50% of the values of $\overline{X}$ are below 19.1 and 50% are above 19.1.
Therefore, the distribution of $\overline{X}$ is symmetric for this case. Thus, the expected value of $\overline{X}$ must be 19.1.
Therefore, ${\mu}_{\overline{X}}=19.1$.
Referring to a. part, we know that ${\mu}_{\overline{X}}=\mu $. So, the value of $\mu =19.1$.
We are considered that $P\left(\overline{X}>19.1\right)=0.2$. So, we can conclude that 20% values of $\overline{X}$ are above 19.1 and 80% of values are below 19.1.
Therefore, the distribution is not symmetric for this case. For this to possible, 19.1 will fall to the right of the mean of the distribution.
Therefore,
$\begin{array}{l}{\mu}_{\overline{X}}<19.1\\ i.e\text{\hspace{0.33em}}\mu <19.1\end{array}$
So,$\mu $ is less than 19.1 years.
Surface roughness of pipe. Refer to the Anti-CorrosionMethods and Materials (Vol. 50, 2003) study of the surface roughness of oil field pipes, Exercise 2.46 (p. 96). Recall that a scanning probe instrument was used to measure thesurface roughness x (in micrometers) of 20 sampled sectionsof coated interior pipe. Consider the sample mean,$\overline{\mathbf{X}}$.
1.72 | 2.50 | 2.16 | 2.13 | 1.06 | 2.24 | 2.31 | 2.03 | 1.09 | 1.40 |
2.57 | 2.64 | 1.26 | 2.05 | 1.19 | 2.13 | 1.27 | 1.51 | 2.41 | 1.95 |
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