Aset of data has a mean of 56.1. the data follows a normal distribution curve and has a standard deviation of 8.2. find the probability that a randomly selected value is greater than 67.5 a. 0.0823 b. 1.39 c. 0.9177 d. –1.08

A.  0.0823

Step-by-step explanation:

Given : Mean =56.1

Standard deviation =8.2

P(x>67.5) will be found as follows:

The z-score is given by:

Substitute the values of means ans standard deviation in it, we get

As

P(z<1.39)=0.9177

thus:

P(X>67.5)=1-P(z<1.39)

⇒P(X>67.5)=1-0.9177

∴P(X>67.5)=0.0823

step-by-step explanation:

x^2 -12x + 36 + y^2 = 36

x^2 + y^2 = 12x

r^2 = 12rcos(theta)

r = 12 cos(theta)

i believe the answer is letter c) 8

step-by-step explanation: