On a dry surface, the braking distance (in meters) of a certain car is a normal

Question

On a dry surface, the braking distance (in meters) of a certain car is a normal distribution with mean 45.1 m and standard deviation 0.5. Find the braking distance that corresponds to z = 1.8 Find the braking distance that represents the 91st percentile. Find the z-score for a braking distance of 46.1 m. Find the probability that the braking distance is less than or equal to 45 m. Find the probability that the braking distance is greater than 46.8 m. Find the probability that the braking distance is between 45 m and 46.8 m.

Explanation / Answer

mean is 45.1 and sd is 0.5

a) when z is 1.8, (x-45.1)/0.5=1.8 or x=1.8*0.5+45.1=46

b) for 0.91 , the z value from normal distribution table is 1.35, thus x=1.35*0.5+45.1=45.775

c) z is (46.1-45.1)/0.5=2

d) P(x>46.8)=P(z>(46.8-45.1)/0.5) or P(z>3.4) or answer is 0

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