The health food store that you manage employs 7 clerks. Individual hourly wages are $8.10, $8.25, $8.45, $8.75, $9.00, $9.25, an
2 answers:
Answer:
16,848+17,160+17,576+18,200+18,720+19,240+21,840=
129,584
Step-by-step explanation:
Annually= Yearly, A year has 365 days.
They work 40 hours a week, 8*5=40.
There are aprox, 10 holidays in a year. (You can change this in your work)
They have 2 weeks unpaid vacation, 14 days.
There are aprox, 52 weeks in a year. (You can change this in your work)
52 weeks - 2 weeks of unpaid vacation= 50 weeks.
Now Let's calculate the weekly income, for
$8.10*40=$324 per week.
$8.25*40=$330 per week.
$8.45*40=$338 per week.
$8.75*40=$350 per week.
$9.00*40=$360 per week.
$9.25*40=$370 per week.
$10.50*40=$420 per week.
Now Let's calculate the annually income, for each clerk.
324*50=16,200 + 648=16,848
330*50=16,500 + 660=17,160
338*50=16,900 + 676=17,576
350*50=17,500 + 700=18,200
360*50=18,000 + 720=18,720
370*50=18,500 + 740=19,240
420*50=21,000 + 840=21,840
Now the total annual payroll=
16,848+17,160+17,576+18,200+18,720+19,240+21,840=
129,584
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Answer:
The 90th percentile for the distribution of the total contributions is $6,342,525.
Step-by-step explanation:
To solve this question, we need to understand the normal probability distribution and the central limit theorem.
Normal probability distribution
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean and standard deviation
, the zscore of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
Central Limit Theorem
The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean and standard deviation
, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
and standard deviation
.
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
For sums of size n, the mean is and the standard deviation is
In this question:
The 90th percentile for the distribution of the total contributions
This is X when Z has a pvalue of 0.9. So it is X when Z = 1.28. Then
By the Central Limit Theorem
The 90th percentile for the distribution of the total contributions is $6,342,525.