Suppose a normal distribution has a mean of 222 and a standard deviation of 16. What is the probability that a data value is between 206 and 230? Round
your answer to the nearest tenth of a percent.
A. 91.0%
B. 66.9%
C. 53.3%
D. 84.0%

C) 53.3%

The probability that a data value is between 206 and 230

P( 206 ≤X≤230) = 0.5328 = 53.3%

Step-by-step explanation:

Explanation

Given that Mean of the Normal distribution(μ)  = 222

Given that the standard deviation of the Normal distribution (σ) = 16

Let 'X' be the random variable in the Normal distribution

we have to find that the probability that a data value is between 206 and 230

solution:-

Step(i):-

Let  'X' = 206

Let X = 230

Step(ii):-

The probability that a data value is between 206 and 230

P( 206 ≤X≤230) = P( -1≤Z≤0.5)

                          = |A(0.5)+A(-1)|

                         = 0.1915+0.3413

                         = 0.5328

final -

The probability that a data value is between 206 and 230

P( 206 ≤X≤230) = 0.5328 = 53.3%

hope this .

r3t40

correct choice is b

step-by-step explanation:

when you calculate the average rate of change of a function, you are finding the slope of the secant line between the two points. for the function the average rate of change from point to point can be calculated as

to find the average rate of change from day 2 to day 10, note that at day 2, 8 houses were sold, so a(2,8) and at day 10, 2 houses were sold, so b(10,2). thus,