Question

Salary of a travel management professional. According to the most recent Global Business Travel Association (GBTA) survey, the average base salary of a U.S. travel management professional is \(94,000. Assume that the standard deviation of such salaries is \)30,000. Consider a random sample of 50 travel management professionals and let $\overline{\mathbf{\chi }}$ represent the mean salary for the sample.

1. What is${\mathbf{\mu }}_{\overline{\mathbf{\chi }}}\mathbf{?}$
2. What is${\mathbf{\sigma }}_{\overline{\mathbf{\chi }}}\mathbf{?}$
3. Describe the shape of the sampling distribution of$\overline{\mathbf{\chi }}.$
4. Find the z-score for the value$\overline{\mathbf{\chi }}\mathbf{=}\mathbf{86}\mathbf{,}\mathbf{660}$
5. Find$\mathbf{P}\left(\overline{\mathbf{\chi }}\mathbf{>}\mathbf{86}\mathbf{,}\mathbf{660}\right).$

Short answer

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Step-by-step solution

(a) The data is given below

The calculation is given below:

Given,

$\begin{array}{l}\mu =94,000\\ \sigma =30,000\\ \mathrm{n}=50\end{array}$

$\begin{array}{c}{\mu }_{\overline{\mathrm{\chi }}}={\mu }_0=94000\\ {\mu }_{\overline{\mathrm{\chi }}}=94000\end{array}$

(b) The data is given below

The calculation is given below:

$\begin{array}{c}{\sigma }_{\overline{\mathrm{\chi }}}=\frac{\sigma }{\sqrt{\mathrm{n}}}\\ =\frac{30,000}{\sqrt{50}}\\ =4242.64068712\\ \simeq 4242.6407\end{array}$

${\sigma }_{\overline{\mathrm{\chi }}}=4242.6407$

(c) The data is given below

The calculation is given below:

As here $\mathrm{n}\ge 30$

As a result, the sample size is sufficient to follow the normal distributions. As a result, the distribution's form is normal.

The chart is given below:

(d) The data is given below

The calculation is given below:

$\mathrm{Zscore}=\frac{\mathrm{\chi }-\mu }{\sigma /\sqrt{\mathrm{n}}}$

$\begin{array}{l}\mathrm{\chi }=86660\\ \mu =94000\\ \mathrm{n}=50\end{array}$

$\begin{array}{c}\mathrm{Zscore}=\frac{86660-94000}{4242.6407}\\ =-1.73\end{array}$

(e) The data is given below

The calculation is given below:

$\begin{array}{c}\mathrm{P}\left(\overline{\mathrm{\chi }}>8660\right)=1-\mathrm{P}\left(\overline{\mathrm{\chi }}<8660\right)\\ =1-\mathrm{P}\left(\frac{\overline{\mathrm{\chi }}-{\mu }_{\overline{\mathrm{\chi }}}}{{\sigma }_{\overline{\mathrm{\chi }}}}<\frac{8660-94000}{4242.6407}\right)\\ =1-\mathrm{P}\left(\mathrm{Z}<-1.73\right)\end{array}$

Standard normal table:

$\begin{array}{l}=1-0.9581848\\ =0.0418152\\ \simeq 0.0418\\ \mathrm{P}\left(\overline{\mathrm{\chi }}>86660\right)=0.0418\end{array}$