The compressive strength of samples of cement can be modeled by a normal distribution with a mean of 6000 kilograms per square centimeter and a standard deviation of 100 kilograms per square centimeter. a. What is the probability that a sample’s strength is less than 6250 Kg/cm²? b. What is the probability that a sample’s strength is between 5800 and 5900 Kg/cm²? c. What strength is exceeded by 95% of the samples?

The normal distribution, also known as the Gaussian distribution, is the most important probability distribution in statistics for independent, random variables.

The compressive strength of samples of cement can be modeled by a normal distribution with a mean of 6000 kilograms per square centimeter and a standard deviation of 100 kilograms per square centimeter.

Let x1, x2, x3.....xn be the compressive strength of cement in a sample of size n.

X follows N(6000, 100).

Then we introduce standard normal variate Z.

Z = [tex]\frac{X - 6000}{100}[/tex]

P( X <6250) = [tex]P(\frac{X - 6000}{100} < \frac{6250 - 6000}{100}) = P(Z < 2.5) = 0.9937[/tex] (using excel function NORM.S.DIST(2.5,TRUE))

P( 5800 < X < 5900) =

[tex]P(\frac{5800 - 6000}{100} < \frac{X - 6000}{100} < \frac{5900 - 6000}{100}) = P(-2 < Z < -1) = 0.1359[/tex]

(using excel function NORM.S.DIST(-1,TRUE) - NORM.S.DIST(-2,TRUE))

P(X > x) = 0.95

[tex]P(\frac{X - 6000}{100} > \frac{x- 6000}{100}) = 0.95\\\\P(Z > \frac{x- 6000}{100}) = 0.95\\\\1 - P(Z < \frac{x- 6000}{100}) = 0.95\\\\P(Z < \frac{x- 6000}{100}) = 0.05\\\\Z = -1.644853627\\\\\frac{X-6000}{100} = -1.644853627\\\\X = 5835.514637[/tex]

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