Determine if 0.669429565254154 is rational or irrational and give a reason for your answer

The weekly salary paid to employees of a small company that supplies part-time laborers averages $750 with a standard deviatio

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Answer:

(a) The fraction of employees is 0.84.

(b)

$\mu=850\\\\\sigma=450$

(c)

$\mu=787.5\\\\\sigma=472.5$

(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

$z=\frac{X-\mu}{\sigma}= \frac{300-750}{450}=\frac{-450}{450}=-1\\\\P(X>300)=P(z>-1)=0.84$

(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.

$\mu_{new}=E(x+C)=E(x)+E(C)=\mu+C=750+100=850$

The standard deviation stays the same ($450)

$\sigma_{new}=\sqrt{\frac{1}{N} \sum{[(x+C)-(\mu+C)]^2} } =\sqrt{\frac{1}{N} \sum{(x+C-\mu-C)^2} }\\\\ \sigma_{new}=\sqrt{\frac{1}{N} \sum{(x-\mu)^2} } =\sigma$

(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.

$\mu_{new}=E(ax)=a*E(x)=a*\mu=1.05*750=487.5$

The standard deviation increases by a 5% ($472.5)

$\sigma_{new}=\sqrt{\frac{1}{N} \sum{[(ax)-(a\mu)]^2} } =\sqrt{\frac{1}{N} \sum{a^2(x-\mu)^2} }\\\\ \sigma_{new}=\sqrt{a^2}\sqrt{\frac{1}{N} \sum{(x-\mu)^2}}=a*\sigma=1.05*450=472.5$