Answer:
The travel time that separates the top 2.5% of the travel times from the rest is of 91.76 seconds.
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.
Mean of 80 seconds and a standard deviation of 6 seconds.
This means that
What travel time separates the top 2.5% of the travel times from the rest?
This is the 100 - 2.5 = 97.5th percentile, which is X when Z has a p-value of 0.975, so X when Z = 1.96.
The travel time that separates the top 2.5% of the travel times from the rest is of 91.76 seconds.
Answer:
Step-by-step explanation:
c
S(20)=
5*20-2=100-2=98
The answer is 4/15, because we have to get the denominaters to be the same, so the LCM of 5 and 15 is 15, so 15/5 is 3 and 4*3 is 12. Now, its 12/15-8/15 which is 4/15. Hope this helps!
Answer:
its a estamite just guess
Step-by-step explanation:
Answer:
i worked out the equation provided and all i get is (y = -2.5)
Step-by-step explanation: