et X represent the full height of a certain species of tree. Assume that X has a normal distribution with a mean of 191.8 ft and a standard deviation of 4 ft. A tree of this type grows in my backyard, and it stands 182.2 feet tall. Find the probability that the height of a randomly selected tree is as tall as mine or shorter.

The probability that the height of a randomly selected tree is as tall as mine or shorter.

P(X≤182.2) = 0.0082

Step-by-step explanation:

Step(i):-

Mean of the Population = 191.8 ft

Standard deviation of the Population = 4 ft

Let 'X' be the random variable in normal distribution

X = 182.2 feet

Step(ii):-

The probability that the height of a randomly selected tree is as tall as mine or shorter.

P(X≤182.2) = P( z≤ -2.4)

                =  1-P(z>2.4)

                = 1-(0.5 + A(2.4))

              =  0.5 - A(2.4)

             = 0.5 - 0.4918

            = 0.0082

The probability that the height of a randomly selected tree is as tall as mine or shorter.

P(X≤182.2) = 0.0082

x+y=90

x=30-y=x

step-by-step explanation:

if you understand you will get the answer

because 1.2 isnt half of 6 duh