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Burka [1]
1 year 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]1 year 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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Step-by-step explanation:

We are given that the average annual amount American households spend on daily transportation is $6312 (Money, August 2001). Assume that the amount spent is normally distributed.

(a) It is stated that 5% of American households spend less than $1000 for daily transportation.

Let X = <u><em>the amount spent on daily transportation</em></u>

The z-score probability distribution for the normal distribution is given by;

                          Z  =  \frac{X-\mu}{\sigma}  ~ N(0,1)

where, \mu = average annual amount American households spend on daily transportation = $6,312

           \sigma = standard deviation

Now, 5% of American households spend less than $1000 on daily transportation means that;

                      P(X < $1,000) = 0.05

                      P( \frac{X-\mu}{\sigma} < \frac{\$1000-\$6312}{\sigma} ) = 0.05

                      P(Z < \frac{\$1000-\$6312}{\sigma} ) = 0.05

In the z-table, the critical value of z which represents the area of below 5% is given as -1.645, this means;

                           \frac{\$1000-\$6312}{\sigma}=-1.645                

                            \sigma=\frac{-\$5312}{-1.645}  = 3229.18

So, the standard deviation of the amount spent is $3229.18.

(b) The probability that a household spends between $4000 and $6000 is given by = P($4000 < X < $6000)

      P($4000 < X < $6000) = P(X < $6000) - P(X \leq $4000)

 P(X < $6000) = P( \frac{X-\mu}{\sigma} < \frac{\$6000-\$6312}{\$3229.18} ) = P(Z < -0.09) = 1 - P(Z \leq 0.09)

                                                            = 1 - 0.5359 = 0.4641

 P(X \leq $4000) = P( \frac{X-\mu}{\sigma} \leq \frac{\$4000-\$6312}{\$3229.18} ) = P(Z \leq -0.72) = 1 - P(Z < 0.72)

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(c) The range of spending for 3% of households with the highest daily transportation cost is given by;

                    P(X > x) = 0.03   {where x is the required range}

                    P( \frac{X-\mu}{\sigma} > \frac{x-\$6312}{3229.18} ) = 0.03

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                         {x-\$6312}=1.88\times 3229.18  

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