The heights of the trees in a forest are normally distributed, with a mean of 25 meters and a standard deviation of 6 meters. What is the probability that a randomly selected tree in the forest has a height LESS THAN OR EQUAL TO 37 meters? Use the portion of the standard normal table given to help answer the question.

A. 98.6%

B. 2.3%

C. 97.7%

D. 1.4%

Thanks for the supplied normal table; very handy.

We have random variable H with μ=25, σ=6 and we seek P(H < 37).

H=37 corresponds to a z score  (number of standard deviations from the mean) of

z = (H - μ)/σ = (37 - 25)/6 = 2

We just found out

P(H < 37) = P(z < 2)

which we can look up.  Of course for a nice round z score of 2 we don't really need to look that one up.  By the 68-95-99.7 rule, two sigma corresponds to 95%, i.e.

P(-2 < z < 2) = 95%

That means

P(0 < z < 2) = 95/2 = 47.5 %

Of course

P(z < 0) = 50%

so

P(z < 2) = P(z < 0) + P(0 < z < 2) = 50 + 47.5 = 97.5%

C

For completeness, let's look up z=2 in the table.  It doesn't say but this particular table is telling us the answer we want directly.

P(z < 2) = .9772 = 97.72%

which is a bit more accurate than the rule, but in stats that doesn't usually matter much.

3/6 chance

step-by-step explanation:

half of the nubers are odd numbers (1,3,5) which in that case if you would be told to simplify it would be 1/2

27/4x^6 y^8

step-by-step explanation:

4(3x^2y^4)^3 (2x^3y^5)^4

=

108x^6 y^12 /16x^12 y^20

=

108 /16x^6 y^8

=

27 /4x^6 y^8