Answer:
0.281 ; 0.864
Step-by-step explanation:
1.)
p decreases as n increases (n increases, p--> 0)
λ ≥ 0
P(x = x) = (e^-λ * λ^x) ÷ x!
λ for Ford = 1.27
B) zero problems :
P(x = 0) :
(e^-1.27 * 1.27^0) ÷ 0! = 0.281
P(x = 0) = 0.281
Two or fewer problems :
P(x ≤ 2) = p(x = 0) + p(x = 1) + p(x = 2)
P(x ≤ 2) = 0.281 + 0.357 + 0.226
P(x ≤ 2) = 0.864
D.) It allows for early detection and thus adequate curtailment of the problem at an a stage in which it seems easier to curb.
Answer:
Decay
Step-by-step explanation:
Answer:
0.6826 = 68.26% probability Z will be within one standard deviation of average.
Step-by-step explanation:
Normal Probability Distribution
Problems of normal distributions can be solved using the z-score formula.
In a set with mean
and standard deviation
, the z-score of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.
What is the probability Z will be within one standard deviation of average?
This is the p-value of Z = 1 subtracted by the p-value of Z = -1.
Z = 1 has a p-value of 0.8413.
Z = -1 has a p-value of 0.1587.
0.8413 - 0.1587 = 0.6826
0.6826 = 68.26% probability Z will be within one standard deviation of average.
E = mc^2......divide both sides by c^2
E / c^2 = m <===
Answer:
Step-by-step explanation:
3.5805