Answer:
a) 1
b) X = 34.313
c) 4.79*10⁻⁷
d) 36.67449
Step-by-step explanation:
a) Given data
Mean: μ = 37
n = 16
Standard deviation: σ = 10
For a group of 16 individuals, probability that the average percent of fat calories consumed is more than 5 is given by
P(x ≥ 5)
Changing into standard normal variate
P(z ≥ (5 - 36)/(10/√16)) = P(z ≥ -12.4)
P(z ≥ -12.4) = 1 - P(z < -12.4) = 1 - 1.3*10⁻³⁵ ≈ 1
From excel NORM.ES.DIST(-12.4, TRUE): P(z < -12.4) = 1.3*10⁻³⁵
b) First quartile for the average percent of fat calories
Finding the value of Z at P = 0.25
Z(P = 0.25) = -0.67449 ( From Excel =NORM.S.INV(0.25))
Z = (X - μ)/(σ/√n)
⇒ -0.67449 = (X - 36)/(10/√16)
⇒ X = 34.313
c) Given
n = 100
P(85 < x < 1250)
Z₁ = (X₁ - μ)/σ ⇒ Z₁ =(85 - 36)/10 = 4.9
Z₂ = (X₂ - μ)/σ ⇒ Z₂ =(1250 - 36)/10 = 121.4
P(4.9 < Z < 121.4) = P(Z < 121.4) - P(Z ≤ 4.9) = = 4.79*10⁻⁷
d) n = 100
Third quartile for the average percent of fat calories
Finding the value of Z at P = 0.75
Z(P = 0.75) = 0.67449 ( From Excel =NORM.S.INV(0.75))
Z = (X - μ)/(σ/√n)
⇒ 0.67449 = (X - 36)/(10/√100)
⇒ X = 36.67449
