Question

Suppose that we do not know the mean and the standard deviation, and that we have calculated the sample mean and that we have calculated the sample mean and sample standard deviation of the compressive strength based on 10 samples to be 6000kg/cm^2 and 100 kg/cm^2.

(a) What is the probability that a sample’s strength is greater than 5800 Kg/cm2?

(b) What is the probability that a sample’s strength is between 5800 Kg/cm2 and 5950 Kg/cm2?

(c) What strength is exceeded by 85% of the samples?

Answer 1

P(X < A) = P(Z < (A - mean)/standard deviation)

a) P(sample strength is greater than 5800) = P(X > 5800)

= 1 - P(X < 5800)

= 1 - P(Z < (5800 - 6000)/100)

= 1 - P(Z < -2)

= 1 - 0.0228

= 0.9772

b) P(5800 < X < 5950)

= P(X < 5950) - P(X < 5800)

= P(Z < (5950 - 6000)/100) - P(Z < -2)

= P(Z < -0.5) - P(Z < -2)

= 0.3085 - 0.0228

= 0.2857

c) Let K be the strength exceeded by 85% of sample

P(X > K) = 0.85

P(X < K) = 1 - 0.85 = 0.15

P(Z < (K - 6000)/100) = 0.15

From standard normal distribution table, take Z value corresponding to probability of 0.15

(K - 6000)/100 = -1.04

K = 5896

5896 kg/cm² is the strength is exceeded by 85% of the samples