# QSO

QSO 510: Module 2 Homework
Notes:

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Problem 1
If many samples of size 15 (that is, each sample consists of 15 items) were taken from a large normal population with a mean of 18 and a variance of 5, what would be the mean, variance, standard deviation, and shape of the distribution of sample means? Give reasons for your answers. Note: Variance is the square of the standard deviation. Adapted from Statistics for Management and Economics by Watson, Billingsley, Croft, and Huntsberger. Fifth Edition. Chapter 7 Page 308. Allyn and Bacon. 1993x = 18  2x=5 shape = normal
Therefore x= 2x = 5 = 2.2360
If a large number of samples, each of size 15, are selected from the population and a sample mean is selected computed for each sample, the mean, variance, and shape of the distribution of sample means would be as follows: x= 18. The standard deviation of the sample means x can be computed as given below: x = X = 2.2360 = 0.5773 n 15.  Variance is the square of the standard deviation. Therefore, the variance of the sample means ²x can be computed as follows:
²x = ²x = 5 = 0.3333 n 15.
The shape of the distribution of sample means = normal since sample size n 30.
Problem 2
If many samples of size 100 (that is, each sample consists of 100 items) were taken from a large non-normal population with a mean of 10 and a variance of 16, what would be the mean, variance, standard deviation, and shape of the distribution of sample means? Give reasons for your answers. Note: Variance is the square of the standard deviation. Adapted from Statistics for Management and Economics by Watson, Billingsley, Croft, and Huntsberger. Fifth Edition. Chapter 7 Page 308. Allyn and Bacon.
1993 x = 10 2x=16 shape =non- normal
Therefore x= 2x = 16 = 4.
If a large number of samples, each of size 100, are selected from the population and a sample mean is selected computed for each sample, the mean, variance, and shape of the distribution of sample means would be as follows: x= 10
The standard deviation of sample means x can be computed as given below:
x = X = 4 = 0.4 n 100.
Variance is the square of the standard deviation. Therefore, the variance of the sample means ?²x? can be computed as follows:
²x = ²x = 16 = 0.16 n 100.
The shape of the distribution of sample means = normal since sample size n 30.
Problem 3
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.

a) If you pick an employee at random, what is the probability that the number of absences for this one employee would exceed 25 days?
b) If many samples of 36 employees each are taken and sample means computed, 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.
c) 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 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?

Source: Statistics for Management and Economics by Watson, Billingsley, Croft and Huntsberger. Chapter 7 Page 305. Fifth Edition. Allyn and Bacon 1993.

a) For x = 25, z = 25-21/10 = 0.4 From the z table p (0 < z < .4) = .1554

Therefore, p (x > .4) = .5 – p(0 < z < .4) = .5 – .1554 = .3446 = 34.46%

b) x = 21 x=10
If a large number of samples, each of size 36, are selected from the population and a sample mean is selected computed for each sample, the mean, variance and shape of the distribution of sample means would be as follows: x= 21.

The standard deviation of sample means x can be computed as given below:
x = X = 10 = 1.6666 n 36.
Variance is the square of the standard deviation. Therefore, the variance of the sample means ²x can be computed as follows:
²x = ²x = 10² = 2.7777 n 36.
The shape of the distribution of sample means = normal since sample size n 30.

c) This question concerns the distributions of sample means. For the distribution of sample means: x = 21 and x = X = 10 = 1.6666 n 36, and the shape would be normal.

For x = 25 z = 25 – 21 = 2.4, 1.6666.  From the z-table P(021) = 0.5 – P(0(3-3.1)/0.4)  =P(z>-0.25) =1-0.4013 (from z table) =0.5987

d) Sample size,n= 64 mean remains same 3.1
standard error of the mean (standard deviation of sample means) expected to be = s/vn = 0.4/v64 = 0.4/8 =0.05 expected shape of the distribution of sample means will be bell shaped as this is normally distributed.

g) If a random sample of 64 customers is selected, the probability that the sample mean would exceed 3 minutes=P(x>3) =P((x-µ)/e >(3-3.1)/0.05) =P(z>-2) =1-0.0228 (from z table) =0.9772.

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