Probabilities Question

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lizzpalmer

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Jun 20, 2011
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NOTE: For these questions we must word-process them and not use a program like excel.

Here is my question:

Time lost due to employee absenteeism is an important problem for many companies. The human resources department of Western Electronics has studied the distribution of time lost due to absenteeism by individual employees. During a one-year period, the department found a mean of 21 days and a standard deviation of 10 days based on data for all the employees.
  1. If you pick an employee at random, what is the probability that the number of absences for this one employee would exceed 25 days?
  2. If many samples of 36 employees each are taken and sample means computed, a distribution of sample means would result. What would be the mean, standard deviation and shape of the distribution of sample means for samples of size 36? Give reasons for your answers.
  3. A group of 36 employees is selected at random to participate in a program that allows a flexible work schedule, which the human resources department hopes will decrease the employee absenteeism in the future. What is the probability that the mean for the sample of 36 employees randomly selected for the study would exceed 25 days?
I'm pretty green at this stuff but what I have gathered is that the mean is 21 and standard deviation is 10. (which is stated in the question)
I'm not sure what formula I would use to solve these questions. Any help getting started would be appreciated!
 
Let's just work on #1. Consider \(\displaystyle \frac{25-21}{10}\)
 
I think I have an idea but not totally. I have one in front of me. If I look at a z score of .4 the first row under 0.00 gives me .1554 but I'm not sure which column I would pick, whether it would be the 0.00 colum or one of the other ones. How do I tell which one to use?
 
Oh I think I've got it. The other rows are say if you had .41 or .42 then you would move over to those rows. Correct?
 
so what we have done so far is:

25 (i'm not sure what to name this number) - mean / standard deviation = z

.4 produces a probability of .1554

He makes us write out the problems with all the symbols to get full credit. z = 25 - Ux /Ox

correct?
 
For x = 25, z =
25-21/10 = .4
From the z table p (0 < z < .4) = .1554
Therefore, p (x > .4) = .5 – p(0 < z < .4) = .5 - .1554 = .3446 = 34.46%

is this correct for part 1?
 
I havent heard back about part one but thought I would try part 2. For the second part of the question, would this be correct?
Mean = 21
Variance = 3.1623
Therefore: Ox = 1.7783

The standard deviation of sample mean = .5623

Variance of sample mean =3.1624

The shape of the sample mean is normal size since n ≥ 30.
 
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