# CSU Week 2 Statistics Questionnaire

User Generated

bfbpubpbyngr2015

Mathematics

Columbia Southern University

## Question Description

I'm working on a statistics question and need an explanation and answer to help me learn.

Problems need to include all required steps and answer(s) for full credit. All answers need to be reduced to lowest terms where possible. If the answer is in %, show two decimal places.

Below are excel functions, which you can use to solve for measures of central tendency and variability. You can use the formulas too; but, they will be time consuming.

EXCEL FUNCTIONS

Measures of Central Tendency

Suppose data are in cells A1 to A10

Mean =AVERAGE(A1:A10)

Median =MEDIAN(A1:A10)

Mode =MODE(A1:A10)

Measures of Variability

Suppose data are in cells A1 to A10

Range =MAX(A1:A10)-MIN(A1:A10)

IQR =QUARTILE.EXC(A1:A10,3)-QUARTILE.EXC(A1:A10,1)

Population Variance σ2 =VAR.P(A1:A10)

Sample Variance s2 =VAR(A1:A10) or VAR.S(A1:A10)

Population Standard Deviation σ =STDEV.P(A1:A10)

Sample Standard Deviation s =STDEV(A1:A10) or STDEV.S(A1:A10)

1. Joe and Mary would like to buy their first home in a new city. Below is the list of prices of 20 homes for sale.
 Prices in US Dollars 140 190 265 115 270 240 250 180 160 200 240 280 175 200 310 195 320 105 385 265
1. Find the mean, median, mode. (2 pts)
2. Which measure of central tendency best represent the data? Why? (2 pts)
3. Are there any outliers? If so, name the outlier. (2 pts)
2. The ages of students in a statistics class are listed below.
 Age of students 22 20 25 23 27 30 18 31 19 45 45 19 36 27 19 26 33 40 35 24
1. Create a stem plot (2 pts)
2. What is the mean age of the student? (2 pts)
3. Jenny likes to have at least a 70% average to pass her statistics class. Her previous four test scores are 52%, 66%, 76% and 71%. What is the minimum score Heidi needs on the final exam to pass the class? (4 pts)
4. Consider the following data and corresponding weights.
 Xi Weight (wi ) 3.1 5 2.3 3 2.8 2 4 4
1. Compute the weighted mean. (2 pts)
2. Compute the mean without weighting. (2 pts)
5. The chart below represents the amount of rainfall for 15 days in northern Ohio in July.
 Rainfall in inches 4.1 4.2 3.5 3.9 4.2 4.4 3.8 3.7 3.5 3.4 4.3 3.2 4.3 4.5 4
1. Complete the frequency table. (2 pts)
 Rainfall in inches Frequency Less than 3.0 3.0-3.4 3.5-3.9 4.0-4.5
2. Find the mean of the raw data. (2 pts)
3. Find the median of the raw data. (2 pts)
4. Find the mode of the raw data. (2 pts)
6. Consider a sample with data values of 27, 25, 20, 15, 30, 34, 28 and 25.
1. What is the minimum? (2 pts)
2. What is the first quartile? (2 pts)
3. What is the median? (2 pts)
4. What is the third quartile? (2 pts)
5. What is the maximum? (2 pts)
7. Create the box plot for the data in #6. Label the five points on the box. (4 pts)
8. Using the data in #6.
1. What is the range? (2 pts)
2. What is the interquartile range (IQR)? (2 pts)
9. The histogram of a quantitative variable is positively skewed. The mean of the variable is 35.
1. Which one of the following is a more likely value of the median? (2 pts)
55
30
35
10. A histogram of a given data shows three clear peaks. This may mean that (4 pts)
1. There are three distinct groups in the data
2. There are three modes in the data
3. Both a and b are correct

1 Page
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View attached explanation and answer. Let me know if you have any questions.

01.
1.
Measure
mean
median
modes

Formula used
value
224.25 =AVERAGE (B4:B23)
=MEDIAN (B4:B23)
220
=MODE (B4:B23)
200
240
265

2. Mean is the best representative measure since the shape of the distribution can be approximated to a
symmetric distribution.
3.
Q1 = 178.75 (=QUARTILE (B4:B23, 1)
Q3 = 266.25 (=QUARTILE (B4:B23, 3)
Lower bound and upper bound for quartiles are calculated in Excel file.
Since the all values are between the lower bound (-26.25) and upper bound (516.25), there are no any
outliers for this data set.

02.
1.
Stem
1
2
3
4

Leaf
8,9,9,9
...

Nfvevrat (2410)
Carnegie Mellon University

### Review

Anonymous
I was having a hard time with this subject, and this was a great help.

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