Question

High-strength concrete is supposed to have a compressive strength greater than 6,000 pounds per square inch (psi). A certain type of concrete has a mean compressive strength of 7,000 psi, but due to variability in the mixing process it has a standard deviation of 420 psi, assuming a normal distribution.

What is the probability that a given pour of concrete from this mixture will fail to meet the high-strength criterion?

Answer 1

Answer:

Probability that a given pour of concrete from this mixture will fail to meet the high-strength criterion is 0.00866.

Step-by-step explanation:

We are given that High-strength concrete is supposed to have a compressive strength greater than 6,000 pounds per square inch (psi).

A certain type of concrete has a mean compressive strength of 7,000 psi, but due to variability in the mixing process it has a standard deviation of 420 psi, assuming a normal distribution.

Let X = certain type of concrete compressive strength

SO, X ~ Normal($\mu=7,000 ,\sigma^{2} =420^{2}$)

The z score probability distribution for normal distribution is given by;

                            Z  =  $\frac{X-\mu}{\sigma}$  ~ N(0,1)

where, $\mu$ = mean compressive strength = 7,000 psi

            $\sigma$ = standard deviation = 420 psi

SO, Probability that a given pour of concrete from this mixture will fail to meet the high-strength criterion is given by = P(X < 6,000 psi)

    P(X < 6,000 psi) = P( $\frac{X-\mu}{\sigma}$ < $\frac{6,000-7,000}{420}$ ) = P(Z < -2.38) = 1 - P(Z $\leq$ 2.38)

                                                                    = 1 - 0.99134 = 0.00866

The above probability is calculated by looking at the value of x = 2.38 in the z table which has an area of 0.99134.

Hence, the probability that a given pour of concrete from this mixture will fail to meet the high-strength criterion is 0.00866.