Answer:
There is 8% (P=0.08) that Frances concludes that the new equipment increases the average daily jewelry production when in fact the new equipment has no effect.
Step-by-step explanation:
We have one-sample z-test with a significance level of 0.08 and a power ot the test of 0.85.
In this test, the null hypothesis will state that the new equipment has the same productivity of the older equipment. The alternative hypothesis is that there is a significative improvement from the use of new equipment.
The probability that Frances concludes that the new equipment increases the average daily jewelry production when in fact the new equipment has no effect is equal to the probability of making a Type I error (rejecting a true null hypothesis).
The probability of making a Type I error is defined by the level of significance, and in this test this value is α=0.08.
Then, there is 8% that Frances concludes that the new equipment increases the average daily jewelry production when in fact the new equipment has no effect.
Answer:
a. P(X ≤ 5) = 0.999
b. P(X > λ+λ) = P(X > 2) = 0.080
Step-by-step explanation:
We model this randome variable with a Poisson distribution, with parameter λ=1.
We have to calculate, using this distribution, P(X ≤ 5).
The probability of k pipeline failures can be calculated with the following equation:
Then, we can calculate P(X ≤ 5) as:
The standard deviation of the Poisson deistribution is equal to its parameter λ=1, so the probability that X exceeds its mean value by more than one standard deviation (X>1+1=2) can be calculated as:
= 6.37 * 10^4 = 637 * 10^2 = 63700
In short, Your Answer would be Option A
Hope this helps!
Since the troop made 10 times her then you multiply 950 by 10 and that will give you 9510
Idk because you did not say yes you did how many sides there were
