Question

The percent of fat calories that a person in America consumes each day is normally distributed with a mean of about 36 and a standard deviation of about 10. Suppose that 16 individuals are randomly chosen.
Let = average percent of fat calories.
SX = sum percent of fat calories
a) For the group of 16, find the probability that the average percent of fat calories consumed is more than 5.
b) Find the first quartile for the average percent of fat calories.
c) For the group of 100, find the probability that the sum percent of fat calories consumed is between 85 and 1250.
d) Find the third quartile for the sum percent of fat calories.

Answer 1

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

Answer 2

Answer:

a) P ( Z > -12.4 ) ≈ 1

b) X = 34.313 % average fats consumed

c) P ( 4.9 < Z > -121.4 ) ≈ 0

d) X = 42.745 % sum of fats consumed

Step-by-step explanation:

Given:-

- A random variable (X) is normally distributed with the parameters mean (u) and standard deviation (s.d).

                                  X ~ N ( 36 , 10^2 )

- A sample size of n = 16 individual were chosen.

Solution:-

a) For the group of 16, find the probability that the average percent of fat calories consumed is more than 5.

- The required probability is: P ( X > 5 )

- We will use Z-statistics..(corrected for sample sizes n < 30) for standardizing to determine the required probability.

Where,

             Z-score = ( X - u ) / (s.d / √n)

                           = ( 5 - 36 ) / (10 / √16)

                           = -12.4

So,        P ( X > 5 ) = P ( Z > -12.4 )

- Using standard Z-table:

            P ( Z > -12.4 ) ≈ 1              

b) Find the first quartile for the average percent of fat calories.

- The first quartile for (X) corresponds to probability that the average percent of fat calories consumed is 0.25.

- So equate  P ( X < x ) = P ( X < z ) = 0.25

- Using standard normal table evaluate Z-score value at P ( z ) = 0.25.

                    Z - score = -0.67449

- Use the standardization formula and solve for X:

                    -0.67449 = ( X - u ) / (s.d / √n)

                    X = -0.67449 * (s.d / √n) + u

                    X = 34.313 % fats consumed

- The sum percent of fat calories consumed has a sample size of n = 100, while the normal distribution parameters remain the same as in previous scenario.

c) For the group of 100, find the probability that the sum percent of fat calories consumed is between 85 and 1250.

- The required probability is: P ( 85 < X > 1250 )

- We will use Z-statistics for standardizing to determine the required probability.

Where,

             Z-score,85 = ( X - u ) / (s.d)

                                = ( 85 - 36 ) / (10)

                                = 4.9

            Z-score,1250 = ( X - u ) / (s.d)

                                = ( 1250 - 36 ) / (10)

                                = 121.4

So,        P ( 85 < X > 1250 ) = P ( 4.9 < Z > -121.4 )

- Using standard Z-table:

            P ( 4.9 < Z > -121.4 ) ≈ 0        

d) Find the third quartile for the sum percent of fat calories.

- The 3rd quartile for (X) corresponds to probability that the sum percent of fat calories consumed is 0.75.

- So equate  P ( X < x ) = P ( X < z ) = 0.75

- Using standard normal table evaluate Z-score value at P ( z ) = 0.75.

                    Z - score = 0.67449

- Use the standardization formula and solve for X:

                    0.67449 = ( X - u ) / (s.d)

                    X = 0.67449 * (s.d) + u = 0.67449 * (10) + 36

                    X = 42.745 % sum of fats consumed