# Essay on Statistics: Standard Deviation and Tim Hortons Mean

Submitted By Debashree_Dasgupta
Words: 594
Pages: 3

Problem 1
A marketing company requires that a student must have an average above 75% in their 2nd year fall term courses to get a Co-op. interview. Suppose the company views some courses as being more important than others and so assigns different weights to them as shown below. Below are grades for a student, would the student get an interview with and without considering the weights?

Course marketing finance stats accounting
O.B.
mark %
65
80
70
92
72
Weight
9
8
7
3
4

X = mean = sum of X = 65+80+70+92+72 = 75.8% >= 75% get in interview N 5

Weighted mean = WiXi = 9(65) + 8(80) + 7(70) + 3(92) + 4(72) = 73.5 %, so no interview Wi 9+8+7+3+4

Problem 2
You are studying the salary of new business student graduates. The Alumni office provided you with the following summary for a sample of business students that graduated in 2011. What is the mean starting salary?

starting salary (in \$1,000)
22 – 26
27 - 31
32 - 36
37 - 41
# of students
3
5
8
4

Group mean = sum of FiMi = 3(24) + 5(29)+8(34) + 4(39) n 3+5+8+4 Problem 3
Consider the Tim Horton case (02Lecture-case.pdf). Using Excel’s Data Analysis → Descriptive Statistics tool on question 1 data set of 30 customers (02Lecture-case.xlsx) you get the following output. Do you feel that the times have a small variability or large variability? What do you think of the distribution of the times? Are there any outliers? Do you feel that Store Managers are doing their job in enforcing the 20min rule?

Time Spent at Tim Hortons
Mean
18.73
Standard Error
1.99
Median
17.5
Mode
7
Standard Deviation
10.92
Sample Variance
119.30
Kurtosis
0.249
Skewness
0.725
Range
44
Minimum
3
Maximum
47
Sum
562
Count
30

Mode is the most common. Positive skewness, means more values to the right
1. Mean is 18.73 and standard deviation is 10.92. Coefficient of variation = std. dev / mean = 10.92/18.73 = 58.3% which is really high implying large variability.
2. Distribution of the times – non symmetrical because the skewness is positive (0.725) , showing moderately skewed to the right. This also implies that there are some high values that would pull the average up.
3. Outliers? Minimum time = 3 minutes, z-score = (3-average (18.73))