Frequency Distributions and Central Dispersion

[Name of the Author]

Table of Contents

Case Assignment1

Case 11

Case 22

Wendy's2

MacDonald's3

Comparison of both the samples3

Case 34

Case 46

References8

Session Long Project (SLP)9

References12

Case Assignment

Case 1

To get the best deal on a CD player, Tom called eight appliance stores and asked the cost of a specific model. The prices he was quoted are listed below:

$ 218 $ 125 $ 381 $ 187 $ 231 $ 213 $ 309 $ 230 Find the Standard deviation

The mean price of CD player is calculated below:

Mean = =

Where,

Xis are the sum of the values of sampleN is the number of observations in the sample

= $ 236.75

The standard deviation of the CD player price is calculated below:

Standard Deviation = =

s = 77.44

It means that although the mean price of CD player is found to be $ 236 the standard deviation is quite large and the mean price deviates with approximately $ 77 across the sample. Tom, most of the times, can get CD player for $ 77 less than the mean price of $ 236 and he can also get a CD player for $ 77 more than the mean price of $236.

Case 2

When investigating times required for drive-through service, the following results (in seconds) were obtained. Find the range, variance, and standard deviation for each of the two samples, then compare the two sets of results.

Wendy's 110 113 133 198 124 120 154 110

MacDonald's 105 116 131 176 118 110 135 137

Wendy's

Mean = = = 132.75Range = 88Variance = = = 910.5Standard Deviation = = = 30.174

Analysis of Wendy's drive through service time brought out the above described results. The mean service time is found to be 132.75 sec. The data in the Wendy's sample is dispersed by 88 sec, and that is the dispersion across the data sample. The standard deviation is calculated at 30.174 sec, implying that Wendy's service time deviates with 30.174 seconds when compared to the mean. The mean only provided a central value of the sample data that Tom collected. Mean was also affected by the extreme values. Simply analyzing the mean would not have given Tom figures for effective decision. Standard deviation has provided more authenticity to the data as it has given Tom a range in which most of the values of the data set lie.

MacDonald's

Mean = = = 128.5Range = 71Variance = = = 505.42Standard Deviation = = = 22.4

Analysis of Macdonald's drive through service time brought out the above described results. The mean service time is found to be 128.5 sec. The data in the Macdonald's sample is dispersed by 71 sec, and that is the dispersion across the data sample. The standard deviation is calculated at 22.4 sec, implying that Macdonald's service time deviates with 22.4 seconds when compared to the mean.

Comparison of both the samples

The comparison of descriptive statistics of both the data sets brought out results that favor Macdonald's in speedy drive through service. The mean service time of Macdonald's, 128.5 sec, is less than the mean service time of Wendy's ...