Answer: 0.02055867
Step-by-step explanation:
Given : , since the alternative hypothesis is two-tailed , so the test is a two tailed test.
Sample size : n= 30
Degree of freedom for t-distribution =
Test statistic= t= -2.43
By using the normal t-distribution table ( or calculator ) , we have
The corresponding P-value for two tailed test = 0.02055867
Answer:
The required probability is 0.4098.
Step-by-step explanation:
Consider the provided information.
According to conditional probability that A occurs , given B has occurred.
Assume that the car lot contains 40 percent BMWs, 25 percent Volvos, and 35 percent Jaguars.
Pr(BMWs)=0.40
Pr(Volvos)=0.25
Pr(Jaguars)=0.35
Of the BMWs, 90 percent have navigation systems, 30 percent of the Volvos have navigation systems, and 50 percent of the Jaguars have navigation systems.
Pr(Navigation | BMWs)=0. 90
Pr(Navigation l Volvos)=0.30
Pr(Navigation l Jaguars)=0.50
Total cars with navigation system = 0.40×0.90+0.25×0.30+0.35×0.50
=0.36+0.075+0.175=0.61
Pr(Not Bmw l Navigation system) =
Hence, the required probability is 0.4098.