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Burka [1]
2 years ago
8

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

n of ​$450. ​(a) If the weekly salaries are normally​ distributed, estimate the fraction of employees that make more than ​$300 per week. ​(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? ​(c) If every employee receives a 5​% salary increase for the next​ year, how does the normal model​ change? ​(d) If the lowest salary is ​$300 and the median salary is ​$525​, does a normal model appear​ appropriate? ​(a) If the weekly salaries are normally​ distributed, the fraction of employees that make more than ​$300 per week is approximately nothing. ​(Type an integer or a​ fraction.)
Mathematics
1 answer:
poizon [28]2 years ago
7 0

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

(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.

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