statistics math (1 Viewer)

[stressedadfff]

hey guys can anyone explain b i have the solution but dont understand why its -2/3 and not just 2/3

 

[stressedadfff]

could someone help with b and c pls

 

[Trebla]

hey guys can anyone explain b i have the solution but dont understand why its -2/3 and not just 2/3

View attachment 32284

P(X > 8) = P((X-6)/3 > 2/3)) = P(Z > 2/3)

Note that we cannot say b = 2/3 because P(Z > b) is not equal to P(Z < b). Therefore, to get P(Z < b) we use the symmetry of the standard normal distribution which tells us that
P(Z > 2/3) = P(Z < -2/3)

Hence b = -2/3.

 

[CM_Tutor]

@stressedadfff:

Let the point where is a tangent to be at .

From the diagram, is in quadrant 4 and so and .

Now, you know that lies on the line, and so you have

​

You also know that the gradient of the curve at is since that is the gradient of its tangent. Using calculus, you can find this gradient at and you also know that lies on the curve.

This should allow you to find that:

​

 

[stressedadfff]

P(X > 8) = P((X-6)/3 > 2/3)) = P(Z > 2/3)

Note that we cannot say b = 2/3 because P(Z > b) is not equal to P(Z < b). Therefore, to get P(Z < b) we use the symmetry of the standard normal distribution which tells us that
P(Z > 2/3) = P(Z < -2/3)

Hence b = -2/3.

wouldnt it be 1-p(z<2/3) the complement?

 

[stressedadfff]

@stressedadfff:

Let the point where is a tangent to be at .

From the diagram, is in quadrant 4 and so and .

Now, you know that lies on the line, and so you have

​

You also know that the gradient of the curve at is since that is the gradient of its tangent. Using calculus, you can find this gradient at and you also know that lies on the curve.

This should allow you to find that:

​

thank you!

 

[Trebla]

wouldnt it be 1-p(z<2/3) the complement?

Whilst it is true that
P(Z > 2/3) = 1 - P(Z < 2/3)
It is completely useless to solve the question, because you cannot directly solve for b.

An analogous problem would be if
x² = 1 - y²
You can't then claim that x = y.