STA 2023 Standard Deviation

Question

Please help! I am so lost!

1425      1125      1128      5433      2242      2283

135        1172      1337      1802      2446      1507

2308      3838      2480      1526      2449      2170

1729      8445      2978      3318      1355      1571

2158      4776      1681      4059      1437      2073

1175      996        4388      3801      2130      2406

1209      3319      2015      1883      1668      1447

3217      2821      3967      1807      1087

4467      3634      2217      1340      2463

2906      698        985        559        2546

 

1. Construct a histogram of this data. Use categories that are 500 wide. DO NOT use Excel or another program to draw the graph for you.

 

 

 

 

 

 

 

 

 

1. Find the mean and median of this data set. Looking at these numbers and the shape of the distribution you drew in part 1, describe the skew (if any) of the data. Which measure of central tendency might better describe the “average” number of words in an inauguration address? Why?

 

 

 

 

 

1. Determine the population standard deviation for the number of words in a presidential inauguration address. Are there any obvious outliers (i.e. data points that are very far away from the mean)? If so, what are they? (We’ll talk more about outliers, and get an official definition, next week.)

 

 

 

 

 

1. Take a random sample of this data where n=10. (You can’t choose the same data point twice.) Find the sample mean  and sample standard deviation s for your sample. How does your sample compare to the population? Was your sample a good predictor of the properties of the whole data set, or not?

Answer 1

I'm sorry I used Excel.

I copied your data, pasted in Excel,

did text to columns, and sorted it.

I colored the data in groups of 500 per the directions

so as to make it easy for you to put them in a histogram.

The largest group is the black group which contains

the median and mean average, so that suggests

normal distribution (bell curve). However, the green

group, as shown below is in a very close 2nd place.

I would be tempted to say it is BI-MODAL.

Here are the stat. parameters:

135 min

8445 max

2342.754386 mean

2130 median

1398.394159 std.dev

So as to get a fair, random sample, we need

to examine the distribution. THere are N=57

statistics broken down per bin as follows:

RED 1/57 =1.75%

BLUE 4/57 = 7%

GREEN 12/57 = 21%

ORANGE 9/57 = 15.8%

BLACK 14/57 = 24.6%

BROWN 4/57 = 7%

PURPLE 7/57 = 12.3%

Maroon 4/57 = 7%

final dark black 1/57 = 1.75%

10% of 10 is 1, so anybody at 10% or

more usually gets put in the sample.

So here's the sampling strategy:

green 3

orange 2

black 4

purple 1

Note that the outliers in Red, maroon,

checkered, and dark black do not get picked,

while we pick an extra stat from the middle.

Yes, it had an affect on the mean and standard deviation.

I have uploaded the spreadsheet under the name stats57.xls