Question

The thickness of a type of veneer is approximately normally distributed with mean 5 mm and standard deviation 0.2 mm.

a) What is the probability the thickness of a randomly selected piece is between 4.7 and 5.25 mm?

b) What is the probability the thickness of a randomly selected piece is greater than 5.3 mm?

c) What thickness is the 45th percentile (45% of the thicknesses are less than that value)?

d) What thicknesses constitute the middle 80% of the distribution?

e) Seven layers of veneer are used to make a sheet of plywood. What is the probability that three of them are thicker than 5.3 mm?

Answer 1

μ = 5

σ = 0.2

Z = X -μ / σ

A) p(4.7 < X < 5.25)

= p( 4.7 - 5/0.2 < z < 5.25 - 5/0.2)

= p( -1.5 < z < 1.25)

= 0.4332 + 0.3944 [standard normal distribution table]

= 0.8276

B) p( x > 5.3)

= p( z > 5.3 - 5/0.2)

= p( z > 1.5)

= 0.5 - 0.4332 [standard normal distribution table]

= 0.0668

C) p(X < A) = 0.45

From standard normal distribution table

At p = 0.5 - 0.45 = 0.05 , z = -0.125 (left side)

Z = A - μ / σ

-0.125 = A - 5/0.2

A = 4.975

So, thickness is 4.975 mm

D)

P( -A < X < A) = 0.8

40% of thickness lies less than mean value and 40% lies greater than mean value

From standard normal distribution table

At p = 0.4 , z = +/- 1.28

X1 = Z1 * σ + μ

= (-1.28 * 0.2) + 5

= 4.744

X2 = Z2 * σ + μ

= (1.28 * 0.2) + 5

= 5.256

Thickness interval for middle 80% of the distribution

= (4.744, 5.256)

E)

μ = 5

σ = 0.2/√7 = 0.075

p( x > 5.3)

= p( z > 5.3 - 5/0.075)

= p( z > 4)

= 0.0001 [from standard normal distribution table]

Binomial distribution

P( X = 3) = 7C3 * (0.0001)^3 * (0.4999)^4

= 3.49 x 10^-11