Question

Each section of fencing is normally distributed with mean length equal to 6 feet and standard deviation 0.3 feet. a) If 10 sections of fencing are installed, end to end, find the mean and standard deviation of the total length in feet. b) What is the mean and standard deviation of the total length in inches?

Answer 1

Consider

X : length of each section of fencing in feet.

Given :

X ~ N(mu = 6, sigma² = 0.3²)

E(X) = mean = 6 feet

SD(X) = standard deviation = 0.3 feet.

a) n= number of sections are installed = 10

Y: Total length (in feet) of 10 sections of fencing.

Y = 10 * X

E(Y) = Mean = E(10 *X) = 10 E(X) = 10 *6 = 60 feet

Var(Y) = Var( 10 *X) = 100 Var(Y) = 100 * 0.3² = 9 feet².

SD(Y) = Sqrt(Var(Y)) = 3 feet

Mean and standard deviation of total length of 10 sections are 60 feet and 3 feet respectively.

b) Consider Z : Total length of 10 section in inches.

since 1 feet = 12 inches

Hence random variable Z is defined as

Z = 12 * Y

from part (a)

E(Z) = mean = E(12 * Y) = 12 * E(Y) = 12 *60 = 720 inches.

Var(Z) = Var( 12 * Y) = 144 * Var(Y) = 144 *9 = 1296 inches ².

SD(Z) = sqrt(1296) = 36 inches.

Hence mean and standard deviation of total length in inches are 720 inches and 36 inches respectively.