Answer:
(a) The fraction of employees is 0.84.
(b)
![\mu=850\\\\\sigma=450](https://tex.z-dn.net/?f=%5Cmu%3D850%5C%5C%5C%5C%5Csigma%3D450)
(c)
![\mu=787.5\\\\\sigma=472.5](https://tex.z-dn.net/?f=%5Cmu%3D787.5%5C%5C%5C%5C%5Csigma%3D472.5)
(d) No. The left part of the distribution would be truncated too much.
Step-by-step explanation:
(a) If the weekly salaries are normally distributed, estimate the fraction of employees that make more than $300 per week.
We have to calculate the z-value and compute the probability
![z=\frac{X-\mu}{\sigma}= \frac{300-750}{450}=\frac{-450}{450}=-1\\\\P(X>300)=P(z>-1)=0.84](https://tex.z-dn.net/?f=z%3D%5Cfrac%7BX-%5Cmu%7D%7B%5Csigma%7D%3D%20%5Cfrac%7B300-750%7D%7B450%7D%3D%5Cfrac%7B-450%7D%7B450%7D%3D-1%5C%5C%5C%5CP%28X%3E300%29%3DP%28z%3E-1%29%3D0.84)
(b) If every employee receives a year-end bonus that adds $100 to the paycheck in the final week, how does this change the normal model for that week?
The mean of the salaries grows $100.
![\mu_{new}=E(x+C)=E(x)+E(C)=\mu+C=750+100=850](https://tex.z-dn.net/?f=%5Cmu_%7Bnew%7D%3DE%28x%2BC%29%3DE%28x%29%2BE%28C%29%3D%5Cmu%2BC%3D750%2B100%3D850)
The standard deviation stays the same ($450)
![\sigma_{new}=\sqrt{\frac{1}{N} \sum{[(x+C)-(\mu+C)]^2} } =\sqrt{\frac{1}{N} \sum{(x+C-\mu-C)^2} }\\\\ \sigma_{new}=\sqrt{\frac{1}{N} \sum{(x-\mu)^2} } =\sigma](https://tex.z-dn.net/?f=%5Csigma_%7Bnew%7D%3D%5Csqrt%7B%5Cfrac%7B1%7D%7BN%7D%20%5Csum%7B%5B%28x%2BC%29-%28%5Cmu%2BC%29%5D%5E2%7D%20%20%7D%20%3D%5Csqrt%7B%5Cfrac%7B1%7D%7BN%7D%20%5Csum%7B%28x%2BC-%5Cmu-C%29%5E2%7D%20%20%7D%5C%5C%5C%5C%20%5Csigma_%7Bnew%7D%3D%5Csqrt%7B%5Cfrac%7B1%7D%7BN%7D%20%5Csum%7B%28x-%5Cmu%29%5E2%7D%20%20%7D%20%3D%5Csigma)
(c) If every employee receives a 5% salary increase for the next year, how does the normal model change?
The increases means a salary X is multiplied by 1.05 (1.05X)
The mean of the salaries grows 5%, to $787.5.
![\mu_{new}=E(ax)=a*E(x)=a*\mu=1.05*750=487.5](https://tex.z-dn.net/?f=%5Cmu_%7Bnew%7D%3DE%28ax%29%3Da%2AE%28x%29%3Da%2A%5Cmu%3D1.05%2A750%3D487.5)
The standard deviation increases by a 5% ($472.5)
![\sigma_{new}=\sqrt{\frac{1}{N} \sum{[(ax)-(a\mu)]^2} } =\sqrt{\frac{1}{N} \sum{a^2(x-\mu)^2} }\\\\ \sigma_{new}=\sqrt{a^2}\sqrt{\frac{1}{N} \sum{(x-\mu)^2}}=a*\sigma=1.05*450=472.5](https://tex.z-dn.net/?f=%5Csigma_%7Bnew%7D%3D%5Csqrt%7B%5Cfrac%7B1%7D%7BN%7D%20%5Csum%7B%5B%28ax%29-%28a%5Cmu%29%5D%5E2%7D%20%20%7D%20%3D%5Csqrt%7B%5Cfrac%7B1%7D%7BN%7D%20%5Csum%7Ba%5E2%28x-%5Cmu%29%5E2%7D%20%20%7D%5C%5C%5C%5C%20%5Csigma_%7Bnew%7D%3D%5Csqrt%7Ba%5E2%7D%5Csqrt%7B%5Cfrac%7B1%7D%7BN%7D%20%5Csum%7B%28x-%5Cmu%29%5E2%7D%7D%3Da%2A%5Csigma%3D1.05%2A450%3D472.5)
(d) If the lowest salary is $300 and the median salary is $525, does a normal model appear appropriate?
No. The left part of the distribution would be truncated too much.