A normal distribution has a mean 7 of and a standard deviation of 2. Use the​ 68-95-99.7 rule to find the percentage of values in the distribution between and . What is the percentage of values in the distribution between 7 and 9​?

Explanation:

We have 7 at the center. Going 2 units to the left of center gets us to 7-2 = 5. Going 2 units to the right of center gets us to 7+2 = 9.

Since the mean is 7 and the standard deviation is 2, this tells us that the values 5 and 9 are exactly one standard deviation away from the mean.

Due to the empirical rule (aka 68-95-99.7 rule), we just found that roughly 68% of the population is between 5 and 9.

Half of this is 68/2 = 34%, so this is the percent of values between 7 and 9.

i found two solutions,   n = 2 • ± √3 = ± 3.4641

step-by-step explanation:

y = 5x + 17

step-by-step explanation:

a perpendicular line will be the opposite reciprocal slope of the given equation.   the slope given is -1/5 so the opposite reciprocal is 5

find b:

7 = 5 (-2) + b

7 = -10 + b

17 = b

write equation

y = 5x + 17