Question

The width of a casing for a door is normally distributed with a mean of 24 inches and a standard deviation of 1/8 inch. The width of a door is normally distributed with a mean of 23 and 7/8 inches and a standard deviation of 1/16 inch. Assume independence. Round your answers to three decimal places (e.g. 98.765).

Answer 1

Answer:

a) mean = 0.125, standard deviation = 0.1397

b)0.1867

c) 0.1867

Step-by-step explanation:

The width of a casing for a door is normally distributed with a mean of 24 inches and a standard deviation of 1/8 inch. The width of a door is normally distributed with a mean of 23 7/8 inches and a standard deviation of 1/16 inch. Assume independence. a. Determine the mean and standard deviation of the difference between the width of the casing and the width of the door. b. What is the probability that the width of the casing minus the width of the door exceeds 1/4 inch? c. What is the probability that the door does not fit in the casing?

Let X denote width of a casing for a door and Y be width of a door.If X and Y is normally distributed, X → N(u, σ²) = N(24, (1/8)²)

Also  Y → N(u, σ²) = N(23.875, (1/16)²)

a) Let T be the random variable that denote the difference between width of a casing for a door and width of a door. T = X - Y

E(T) = E(X) - E(Y) = 24 - 23.875 = 0.125

V(T) = V(X) + V(Y) = (1/8)² + (1/16)² = 0.01953

σ(T) = √V(T) = 0.1397

Therefore   T → N(u, σ²) = N(0.125, 0.01953)

b) P(T > 0.25)

Using Z score, $Z=\frac{0.25-0.125}{0.1397} =0.89$

P(T > 0.25) = P(Z > 0.89) = 1 - P(Z<0.89) = 1 - 0.8133= 0.1867

c) P(T < 0)

Using Z score, $Z=\frac{0-0.125}{0.1397} =-0.89$

P(T < 0) = P(Z < -0.89) = 0.1867