The average fuel efficiency of U.S. light vehicles (cars, SUVs, minivans, vans, and light trucks) for 2005 was 21 mpg. If the standard deviation of the population was 2.9 and the gas ratings were normally distributed, what is the probability that the mean mpg for a random sample of 25 light vehicles is Under 20

The probability that the mean mpg for a random sample of 25 light vehicles is 0.042341

Here we used z score formula

So,

For x < 20

= z = 20 - 21/2.9/√25

= z = -1/2.9/5

= z = -1.72414

Now

P-value from Z-Table:

P(x<20) = 0.042341

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0.042341

Step-by-step explanation:

We solve using z score formula

z = (x-μ)/σ/n, where

x is the raw score = 20 mpg

μ is the population mean = 21 mpg

σ is the population standard deviation = 2.9

n = random number of samples

Hence:

For x < 20

= z = 20 - 21/2.9/√25

= z = -1/2.9/5

= z = -1.72414

P-value from Z-Table:

P(x<20) = 0.042341

The probability that the mean mpg for a random sample of 25 light vehicles is Under 20 mpg is 0.042341

(3,0) and (0,6)

Step-by-step explanation:

(x,y) is the only way to make ordered pair.

The x was a 3 so you could cross out the first and last because they put the 3 in the y spot. When is should be (3,0).

Then the y was 6 and that I would be (0,6) so the third would also be crossed out because they put the 6 in the y spot.

So the answer is (3,0) and (0,6).