Answer:
So, the probability that a randomly selected Korean driver will take two or more non-home-based trips per day is P=0.32.
Step-by-step explanation:
We know that the number of non-home-based trips per day taken by drivers in Korea was modeled using the Poisson distribution with λ = 1.15.
We have the Poisson formula:
We calculate:
So, the probability that a randomly selected Korean driver will take two or more non-home-based trips per day is P=0.32.
Answer: NO.
Step-by-step explanation:
As per given , we have to test the hypothesis.
∵ is two-tailed , so our test is a two-tailed test.
Also, the standard deviation is known to be 0.8 , so we use z-test.
Test statistic:
, where = Sample mean
= population mean
= Population standard deviation
n= Sample size
Put
n= 110 , we get
P-value for two tailed test = 2P(Z>|z|)
= 2P(Z>|1.31|) = 2(1-P(Z<1.31)) [∵ P(Z>z)=1-P(Z<z)]
=2(1- 0.9049) [By z-table]
=0.1902
Decision : ∵ P-value (0.1902) > Significance level (0.02).
It means we do not reject the null hypothesis.
[When P-values < Significance level then we reject the null hypothesis.]
Conclusion : We do not have sufficient evidence at the 0.02 level that the valve does not perform to the specifications.