Suppose you have a dataset { x } = { x 1 , x 2 , x 3 , x 4 } consisting of 4 items. You know that x 1 = 25 and x 2 = − 15 and that after standardization ^ x 1 = 0 and ^ x 2 = − 1 . Find mean { x } and std { x } Find x 3 and x 4 given that x 3 ≤ x 4

a) μ = 25,σ = 40 b) x₃ = 0.27864, x₄ = 89.72

Step-by-step explanation:

Given:

{x} = {x₁,x₂,x₃,x₄} -----------equation (1)

where x₁ = 25,x₂ = -15, x₃ ≤ x₄,  ^x₁ = 0, ^x₂=-1

So, equation 1 becomes

{x} = {25,-15,x₃,x₄}

a)

Now, as  ^x₁ = (x₁-μ)/σ

                 0 = (25-μ)/σ

                  μ = 25 = mean{x}

Similarly,   ^x₂=(x₂-μ)/σ

                   -1=(-15-25)/σ

                   σ = 40 =std {x}

b)

As we know that  (x₁+x₂+x₃+x₄)/4=μ

By putting values of x₁,x₂,μ we get,

                  x₃+x₄ = 90 -----------equation (2)

Also we know that

σ²   = (1/4)∑₁⁴²-μ²

40² = 0.25*[25²+(-15)²+x₃²+x₄²]-25²

By solving it we get

x₃²+x₄² = 8050 ----------------equation (3)

Now using equation 2

x₃+x₄ = 90

x₃ = 90-x₄

taking square on both sides

x₃² = 8100 +x₄² -180x₄

bu putting this value in equation 3, we get

8100 + x₄² -180x₄ + x₄² = 8050

-90x₄ + x₄² + 25 = 0

By solving this quadratic equation, we get

x₄ = 89.72 or 0.27864

we will take x₄ = 89.72 as x₃+x₄ = 90 and x₃ ≤ x₄

putting it in equation 2, we get

x₃ = 0.27864

3, 12, 48, 384, 1536, 6144, 24576, 98304

step-by-step explanation:

just keep timing it by 4.