Question

The driver's reaction time in response to a particular potential traffic hazard is the time required from the point of initial detection of the hazard in one’s field of view to the time that vehicle control components are actuated (such as movement of one’s foot to the brake pedal). The following data represents the measurements of the driver’s reaction time in seconds for the male and female participants of the experiment:

0.50	0.49	0.49	0.54	0.44	0.49	0.51	0.46
0.51	0.53	0.51	0.63	0.69	0.76	0.84	0.85

a) find sample mean

sample variance

standard deviation

b) find the median and the quartiles of the sample

median=

q1=

q3=

c) construct a box plot

IQR=

number of outliers=

Answer 1

Number	Values ( X )	Σ ( Xi - X̅ )2
1	0.44	0.0189
2	0.46	0.0138
3	0.49	0.0077
4	0.49	0.0077
5	0.49	0.0077
6	0.5	0.006
7	0.51	0.0046
8	0.51	0.0046
9	0.51	0.0046
10	0.53	0.0023
11	0.54	0.0014
12	0.63	0.0028
13	0.69	0.0127
14	0.76	0.0333
15	0.84	0.0689
16	0.85	0.0743
Total	9.24	0.2713

Part a)

Mean X̅ = Σ Xi / n
X̅ = 9.24 / 16 = 0.5775

Sample Variance = ( (Xi - X̅ )² / n - 1 )

Sample Variance = ( 0.2713 / 16 -1 ) = 0.1345

Sample Standard deviation S_X = √ ( (Xi - X̅ )² / n - 1 )
S_X = √ ( 0.2713 / 16 -1 ) = 0.0181

Part b)

Q1 = (1/4 n + 1/4 )^th value
Q1 = (1/4 * 16 + 1/4 )^th value
Q1 = 0.49

Q2 = (1/2 n + 1/2 )^th value ( Median )
Q2 = (1/2 * 16 + 1/2 )^th value
Q2 = 0.51

Q3 = (3/4 n + 3/4 )^th value
Q3 = (3/4 * 16 + 3/4 )^th value
Q3 = 0.675

Interquartile range = Q3 - Q1
Interquartile range = 0.675 - 0.49
Interquartile range = 0.185

Part c)

Outliers

Low = Q1 - ( 1.5 * IQR ) = 0.415

High = Q3 + ( 1.5 * IQR ) = 0.953

Values lower than 0.415 and higher than 0.953 are outliers

In the data there is no outliers present

number of outliers= 0