Question

The "cold start ignition time" of an automobile engine is being investigated by a gasoline manufacturer. The following times (seconds) were obtained for a test vehicle. 1.68, 1.95, 2.69, 2.36, 3.07, 3.19, 2.57, 1.88

Variance =
Standard Deviation =

Answer 1

Solution:

x	x2
1.68	2.8224
1.95	3.8025
2.69	7.2361
2.36	5.5696
3.07	9.4249
3.19	10.1761
2.57	6.6049
1.88	3.5344
∑x=19.39	∑x2=49.1709

Sample Variance S²=(∑x²-(∑x)²n )n-1

=49.1709-(19.39)287

=49.1709-46.99657

=2.17447

=0.3106

Sample Variance S² = 0.31

Sample Standard deviation S= ( ∑x²-(∑x)²n)n-1

=(49.1709-(19.39)²8)7

=(49.1709-46.9965)7

=2.1744 / 7

=0.3106

=0.5573

Sample Standard deviation S= 0.56