In: Statistics and Probability
Company “Text No Drive” is compiling background information for their new device that senses when texting is going on by the driver and stops the text. The company is interested in using injury statistics for their campaign. They compile information on the cost to the driver (in terms on injury and insurance costs) and also preferences by the driver.
The cost to the driver is described as average cost per injury depending on the amount of texting that the driver admits to doing. It is as follows:
Average cost |
0 |
$800 |
$1,500 |
$3,000 |
Scale : Frequency of texting while driving |
never |
seldom |
several times a week |
every day |
Number of injuries |
7 |
17 |
28 |
43 |
They have piloted a program with the software and it yielded the following preferences for it:
Likert Scale: Preference for this product |
1-Did not like it at all |
2-Liked a few aspects of it |
3-Liked many aspects of it |
4-Liked it very much |
Number of responses |
12 |
24 |
45 |
14 |
(a)
Based on source of data, the type of data is primary data as the company piloted a program to collect data (not secondary type data as not collected from other source or not used for different purpose than that of collection).
Based on nature of data, the type of data is quantitative data as the data directly represents the amount in real numbers. Moreover, values possessing discrete (integral) values only, these are discrete quantitative data.
(b)
Suppose, random number X denotes the amount of average cost in different situation. Also random variable Y denotes the rating obtained in Likert scale.
(1)
Probabilities for each of the average cost categories are as follows.
Average cost (X) | $0 | $800 | $1500 | $3000 | Total |
Scale: Frequency of texting while driving |
never | seldom | several times a week | everyday | - |
Number of injuries | 7 | 17 | 28 | 43 | 95 |
Probability P(X=x) | 7/95=0.074 | 17/95=0.179 | 28/95=0.295 | 43/95=0.452 | 1 |
(2)
Probabilities of type of preference for the product are as follows.
Likert scale (Y): Preference for this product | 1-Did not it at all | 2-Liked a few aspects of it | 3-Liked many aspects of it | 4-Liked it very much | Total |
Number of responses | 12 | 24 | 45 | 14 | 95 |
Probability P(Y=y) | 12/95=0.126 | 24/95=0.253 | 45/95=0.474 | 14/95=0.147 | 1 |
(c)
Probability that average injury cost is at least $800 is given by
(d)
Average injury cost is given by
Now to calculate variance of injury cost we proceed as follows.
So, average injury cost $1941.7 and variance for the injury cost is 1076111.
(e)
Probability that a person will like many aspects of the product = P(Y=3) = 45/95 = 0.474
(f)
Probability that a person will like only few aspects of the product = P(Y=2) = 24/95 = 0.253