The mileages of cars in a fleet of cars are normally distributed with a mean mileage of 52,000 miles and a standard deviation of 2,000 miles. What is the probability of randomly selecting a car with a mileage less than 52,000 miles?

50%

Step-by-step explanation:

We have that the mean is 52000 miles (m) and the standard deviation is 2000 miles (sd), so we are asked:

P (x <52000)

P ((x - m) / sd <(52000 -52000) /2)

P (z <0)

If we look in the normal distribution table we have to:

P (x <52000) = 0.5

That is, the probability is of randomly selecting a car with a mileage less than 52,000 miles 50%

we have three similar triangles, because each has a right angle and shares an angle.   let's write the angles in order: opposite to short leg, long leg, hypotenuse.

cab similar to bad similar to cbd

or as ratios,

ca: ab: cb = ba: ad: bd = cb: bd: cd

we also know

ac = ad + cd

(ad+cd): ab: cb = ba: ad: bd = cb: bd: cd

(ad+cd)/cb=cb/cd

we have cb=6, ad=5 and seek x=cd.

we reject the negative root and conclude x=4

answer: 4

for the free

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