Question

2. The hardness of some cement samples can be modeled by a normal distribution with an average of 6,000 kg / cm2 and a standard deviation of 1000 kg / cm2.

a) What is the probability that the hardness of the sample is less than 6.250 kg / cm2?
b) What is the probability that the hardness of the sample is between 5,800 and 5,900 kg / cm2?
c) Which value is exceeded by 90% of the hardness?
d) Among which values ​​is 95% of the hardness?

Answer 1

a) P(X < 6250)

= P(z < (6250 - 6000)/1000)

= P(z < 0.25)

= 0.5987

b) P(5800 < X < 5900)

= P(-0.2 < z < -0.1)

= 0.0394

c) z score corresponding to 10th percentile = -1.28

Hence,

Score that is exceeded by 90% of the hardness = 6000 - 1.28*1000 = 4720

d) z score corresponding to middle 95% of the area = -1.96, 1.96

Hence,

95% of the hardness lie between (6000 - 1.96*1000) and (6000 + 1.96*1000) i.e. between 4040 kg/ cm2 and 7960 kg/cm2.