SOLUTION: On a dry surface, the braking distance (in meters) of a certain car is a normal distribution with a mean of 45.1 and a standard deviation of 0.5.
a)find the breaking distance that
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-> SOLUTION: On a dry surface, the braking distance (in meters) of a certain car is a normal distribution with a mean of 45.1 and a standard deviation of 0.5.
a)find the breaking distance that
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Question 1137290: On a dry surface, the braking distance (in meters) of a certain car is a normal distribution with a mean of 45.1 and a standard deviation of 0.5.
a)find the breaking distance that corresponds to z=1.8
b)find the breaking distance that represents the 91st percentile
c)find the z-score for a breaking distance of 46.1m
d)find the probability that the breaking distance is less than or equal to 45m
e)find the probability that the breaking distance is greater than 46.8m Answer by Boreal(15235) (Show Source):
You can put this solution on YOUR website! z=(x-mean)/sd
z=1.8=(x-45.1)/0.5
0.9=x-45.1
x=46 m ANSWER a
91st percentile z=1.341
Here, 1.341=(x-45)/0.5
0.67=x-45.1
x=45.8 m (rounded) ANSWER b
less than 45 has a z=(45-45.1)/0.5 or a z<-.1/.5 or z<-0.2
That probability is 0.4207 ANSWER c
greater than 46.8 is z>1.7/0.5 or z> 3.4
This probability is 0.0003
