Suppose a normal distribution has a mean of 38 and a standard deviation of 2. What is the probability that a data value is between 37 and 41? Round your answer to the nearest tenth of a percent.

P ( 37 < x < 41) = P(-0.5 < Z < 1.5) =  0.6247

Step-by-step explanation:

We know mean u = 38  standard dev. s = 2

We want  P ( 37 < x < 41)

so

P( (37 - 38) / 2 <  Z) =  P(-0.5 < Z)  

P( Z <  (41 - 38)/2 ) =  P( Z < 1.5)

Find  P(Z < -0.5) = 0.3085

Find P(Z > 1.5) = 0.0668

so  P(-0.5 < Z < 1.5) =  1  - P(Z < -0.5) - P(Z > 1.5)

P(-0.5 < Z < 1.5) =  1  - 0.3085 -  0.0668

P(-0.5 < Z < 1.5) =  0.6247

P ( 37 < x < 41) = P(-0.5 < Z < 1.5) =  0.6247

what's the question?

step-by-step explanation:

there is no question.

g(x) = 3*(8(3^(x−2))+2)

g(x) = 24 (3^(x-2) )+6

the factor is 2

step-by-step explanation:

let f(x)=8(3^(x−2)+2)

to stretch a graph by 3, it would be 3f(x)

g(x) =3f(x)

g(x) = 3*(8(3^(x−2))+2)

g(x) = 24 (3^(x-2) )+6

x     −2     0     2     4     6

f(x) 1/64 1/16   1/4     1     4

1/64 * z * z = 1/16

i multiplies by z twice because it its 2 steps from -2 to 0

1/64 z^2 = 1/16

multiply by 64 to clear the fraction on the left side

1/4 * 64 z^2 =1/16 *64

z^2 = 4

take the square root of each side

sqrt(z^2) = ±sqrt(4)

z = ±2  

we are multiplying by 2 each time.