Please see the attached file.
15.

A doctor's office staff studied the waiting times for patients who arrive at the office with a request for emergency service. The following data with waiting times in minutes were collected over a one-month period.

2 5 10 12 4 4 5 17 11 8 9 8 12 21 6 8 7 13 18 3

Use classes of 0-4, 5-0 and so in the following:

Class Frequency Cum. Freq Rel. Freq Rel Cum. Freq
0-5 6 6 0.3 0.3
5-10 7 13 0.35 0.65
10-15 4 17 0.2 0.85
15-20 2 19 0.1 0.95
20-25 1 20 0.05 1
Total 20

A. Show the frequency distribution
See above table
B. Show the relative frequency distribution
See above table

c. Show the cumulative frequency distribution.
See above table

d. Show the cumulative relative frequency distribution.
See above table

e. what proportion of patients needing emergency service wait 9 minutes or less?

Number of patients who wait 9 min or less = 12
Total observations = 20
Proportion of patients needing emergency service waits 9 minutes or less
= 12/20 *100
= 60%

Scenario.
3.
A small company was trying to determine the amount of the average order for parts. It had received over 1000 orders for parts. The office manager had the computer select 35 of those orders randomly. He determined that the mean amount of these 35 orders was $389.45 One of the orders was for $350 in miscellaneous cables. He informed his boss that he was 95% certain that the average order that they had received was between $337.97 and $441.03.

1. Determine what is the sample
The randomly selected 35 orders
2. state what is the population
The set of all 1000 orders
3. Identify one or more variables

Amount for parts
Name (Type) of the parts
4. Determine if each variable is quantitative or qualitative and if the scale is nominal, ordinal, interval, or ratio
Amount for parts Quantitative
Name (Type) of the parts Nominal

5. Find one piece of datum

Amount for parts = $350

6. Find a statistic, and
Mean of amount for sample (= $389.45)
7. Find a parameter.
Mean of amount for population

1. Consider a sample with data values of 10, 20, 17, and 16. Compute the mean and median.
Mean = Sum / n = ( 10+20+17+16) / 4 = 63/4 = 15.75

Median
Sorted data
10
16
17
20

Median is the middle value sorted data . Here 16 and 17 are the middle values .Thus median is the average of 16 & 17
Median = 16.5

3.3 13. Consider a sample with data values of 10, 20, 12, 17, and 16. Compute the range and inter quartile range.

Range = Maximum - Minimum

Minimum - 10
Maximum 17
Range = 17-10 = 7

Inter quartile range = 3rd Quartile - 1st Quartile = Q3 -Q1
Where Q1 is the item divide the data in the ratio 25:75
And Q3 is the item divide the data in the ratio 75:25

Sorted Data
10
12
16
17
20

Q3 = 17
Q1 = 12
Inter quartile range = 17-12 =5

17. A home theater in a box is the easiest and cheapest way to provide surround sound for a home entertainment center. A sample of prices is shown here (Consumer Reports Buying Guide, 2004) The prices are for models with a DVD player and for models without a DVD player.
Model with DVD player price ($) Models without a DVD player Price ($)
Sony HT-1800DP 450 Pioneer HTP-230 300
Pioneer HTD-330DV 300 Sony HT-DDW750 300
Sony HT-C800DP 400 Kenwood HTB-306 360
Panasonic SC-HT900 500 RCA RT-2600 290
Panasonic SC-MTI 400 Kenwood HTB-206 300
Model with DVD player price ($) Models without a DVD player Price ($)
Sony HT-1800DP 450 Pioneer HTP-230 300
Pioneer HTD-330DV 300 Sony HT-DDW750 300
Sony HT-C800DP 400 Kenwood HTB-306 360
Panasonic SC-HT900 500 RCA RT-2600 290
Panasonic SC-MTI 400 Kenwood HTB-206 300
Total 2050 Total 1550
Mean 410 Mean 310
Minimum 300 Minimum 290
maximum 500 maximum 360
Range 200 Range 70

a. compute the mean price for models with a dvd player and the mean price for models without a dvd player. What is the additional price paid to have a DVD player included in a home theater unit?

Mean = Total / Number of items

Mean : Model with DVD player = 2050 / 5 = 410
Mean : Model without DVD player = 1550 / 5 = 310

Thus on the average additional $ 100 is needed for model with DVD player

B. compute the range, variance, and standard deviation for the two samples. What does this information tell you about the prices for models with and without a DVD player?

Range = maximum - minimum

Range : Model with DVD player = 500-300 = 200
Range : Model without DVD player = 360-290 = 70

Variance = where Xi is the values of the item and is the mean of the items
Model with DVD player price ($) X X2 Models without a DVD player Price ($) X X2
Sony HT-1800DP 450 202500 Pioneer HTP-230 300 90000
Pioneer HTD-330DV 300 90000 Sony HT-DDW750 300 90000
Sony HT-C800DP 400 160000 Kenwood HTB-306 360 129600
Panasonic SC-HT900 500 250000 RCA RT-2600 290 84100
Panasonic SC-MTI 400 160000 Kenwood HTB-206 300 90000
Total 2050 862500 Total 1550 483700

Variance : Model with DVD player = 862500/5 - 4102 = 5500
Variance: Model without DVD player = 483700/5-3102 = 800

Standard Deviation =
SD : Model with DVD player = 74.1419
SD: Model without DVD player = 28.28427

It can be observed that there is high variability in the price of Model with DVD player

19. The Los Angeles Times regularly reports the air quality index for various areas of Southern California. A sample of air quality index values for provided the following data: 28, 42, 45, 55, 60, 49, and 50.
A compute the range and inter quartile range

Sorted Data
28, 42,45, 49,50,55,60

Range = maximum - minimum = 60-28 =32
Inter quartile range = 3rd Quartile - 1st Quartile = Q3 -Q1
Where Q1 is the item divide the data in the ratio 25:75
and Q3 is the item divide the data in the ratio 75:25

Here Q1 = (42+45)/2 = 43.5
Q3 = (50+55)/2 = 52.5
Inter quartile range = 52.5- 43.5 = 9

B. compute the sample variance and sample standard deviation.

x X2
28 784
42 1764
45 2025
49 2401
50 2500
55 3025
60 3600
329 16099

Mean =
Variance =
SD =

C. A sample of air quality index readings for Anaheim provided a sample mean 48.5, a sample variance of 136, and a sample standard deviation of 11.66. What comparisons can you make between the air quality in Pomona and that in Anaheim on the basis of these
Descriptive statistics.

Coefficient of variation CV =

Pomona : CV= =21.9055

Anaheim : CV=

Since CV for Pomona is low compared to Anaheim , Pomona can be considered as more consistent in air quality index readings