Question

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

Answer 1

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.

Answer 2

Normal distribution has its mean, median and mode coincident on single point. The solutions to the given problems are specified as:

• a) P(X > 300) = 0.8413
• b) The normal model shifts to the right, 100 units, but its structure stays same.
• c) If salary is increased 5%, then the normal model gets scaled .
• d) If lowest salary = $300, and median = $525, then it isn't a normal model as median ≠ mean = $750. It is going to be negatively skewed.

How to get the z scores?

If we've got a normal distribution, then we can convert it to standard normal distribution and its values will give us the z score.

If we have

$X \sim N(\mu, \sigma)$

(X is following normal distribution with mean $\mu$ and standard deviation $\sigma$)

then it can be converted to standard normal distribution as

$Z = \dfrac{X - \mu}{\sigma}, \\\\Z \sim N(0,1)$

(Know the fact that in continuous distribution, probability of a single point is 0, so we can write

$P(Z \leq z) = P(Z < z) )$

Also, know that if we look for Z = z in z tables, the p value we get is

$P(Z \leq z) = \rm p \: value$

For the given case, if we take X = salary of employees weekly of the considered company, then:

$X \sim N(750, 450)$

The  fraction of employees that make more than ​$300 per week is

P(X > 300).

Using the standard normal distribution, we can estimate this  fraction of employees that make more than ​$300 per week as:

$P(X > 300 ) = 1 - P(X \leq 300) = 1 - P(Z = \dfrac{X - \mu}{\sigma} \leq \dfrac{300 - 750}{450} =-1)\\ \\P(X > 300) = 1 - P(Z \leq -1)$

Using the z-table, the p-value for Z = -1 is: 0.1587

Thus, $P(X > 300) = 1 - P(Z \leq -1) = 1 - 0.1587 = 0.8413$

When we add $100 to each value of X, it doesn't change the structure of the graph of normal distribution.
When we increase salary by 5%, it means, new salary $Y = X + 5\% \text{of X} = \dfrac{21X}{20}$

The random variable is scaled by 21/20 = 1.05, so the graph will stay on same origin but it will be stretched a bit in thickness and height.

For the 4th case(d), the median is specified to be $525, but as it was known to us that mean is $750, so mean and median aren't coinciding, and specially, the median is in left of mean, showing that the graph is leaning on right side(negatively skewed). (we deduce it when median < mean, as median shows that mid value is reached, but mean shows that probability is still not reached, so its being late, and reaches later, showing that there is tail in the left of the graph, so being negatively skewed).

Thus, The solutions to the given problems are specified as:

• a) P(X > 300) = 0.8413
• b) The normal model shifts to the right, 100 units, but its structure stays same.
• c) If salary is increased 5%, then the normal model gets scaled .
• d) If lowest salary = $300, and median = $525, then it isn't a normal model as median ≠ mean = $750. It is going to be negatively skewed.

Learn more about standard normal distribution here:

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