Answer:
Probability = 0.35
Step-by-step explanation:
Given :-
- Probability (Highway A Icy) = 0.1
- Probability (Highway B Icy) = 0.15
- Probability (Highway C icy) = 0.15
So :-
- Probability (Highway A not icy) = 1 - 0.1 = 0.9
- Probability (Highway B not icy) = 1 - 0.15 = 0.85
- Probability (Highway C not icy) = 1 - 0.15 = 0.85
Probability of person not getting to work timely
= Probability [Even one of the highway A, B or C is icy]
= 1 - Probability [None of Highway A, B, C is icy]
Since these are independent events, So :-
= Prob. [Highway A not icy & Highway B not icy & Highway C not icy]
= 1 - ( 0.9 x 0.85 x 0.85)
= 1 - 0.65
= 0.35
Answer: 0.1010052
Step-by-step explanation:
Given : Significance level : 
Alternative hypothesis : 
The test statistic value : 
Since , the alternative hypothesis is two-tailed, so the test is a two-tailed test.
Using standard normal distribution table for z, we have
The P-value of two-tailed test will be :-
Hence, the P-value = 0.1010052
It equals approximately 3.74
2 + -9n = 56. -9n = 54. N = -6
Can you describe it a little more- I have no clue how I’m suppose to solve that.
