IQ scores are normally distributed with a mean of 100 and a standard deviation of 15. What is the probability that a randomly chosen person's IQ score will be between 101 and 134, to the nearest thousandth

Step-by-step explanation:

IQ scores are normally distributed with a mean of 100 and a standard deviation of 15. What is the probability that a randomly chosen person's IQ score will be between 101 and 134, to the nearest thousandth

z = (x-μ)/σ, where x is the raw score,

μ is the population mean = 100

σ is the population standard deviation = 15

For x = 101

z = 101 - 100/15

z = 0.066667

Probability value from Z-Table:

P(x = 101) = 0.52658

For x = 134

z = 134 - 100/15

z = 2.26667

Probability value from Z-Table:

P(x= 134) = 0.98829

Therefore,

  b.   the average rate of change of f(x) is less than the average rate of change of g(x).

step-by-step explanation:

since the length of the interval is the same in each case, we only need to compare the change in y-value to see which is greater.

so the change in y-value for function f is (6 )) = 9.

the table tells us

so the change in y-value for function g is (31 -1) = 30.

the change in y-value of f(x) on the interval is less than the change in y-value of g(x) on the same interval, so the average rate of change of f(x) is less than the average rate of change of g(x).