Young's modulus is a quantitative measure of stiffness of an elastic material. Suppose that for metal sheets of a particular type, its mean value and standard deviation are 80 GPa and 1.8 GPa, respectively. Suppose the distribution is normal. (Round your answers to four decimal places.) (a) Calculate P(79 x 81) when n = 25.
(b) How likely is it that the sample mean diameter exceeds 81 when n = 36? You may need to use the appropriate table in the Appendix of Tables to answer this question

(a) Probability that sample mean lies between 79 and 81 is 0.9946.

(b) It is 0.04% likely is it that the sample mean diameter exceeds 81 when n = 36.

Step-by-step explanation:

We are given that metal sheets of a particular type, its mean value and standard deviation are 80 GPA and 1.8 GPA, respectively.

Suppose the distribution is normal.

Let = sample mean diameter

The z-score probability distribution for sample mean is given by;

                                 Z  =    ~ N(0,1)

where, = population mean = 80 GPA

            = standard deviation = 1.8 GPA

            n = sample size = 25

(a) Probability that sample mean lies between 79 and 81 is given by = P(79 < < 81) = P(< 81) - P(79)

       P( < 81) = P( < ) = P(Z < 2.78) = 0.9973

       P( 79) = P( ) = P(Z -2.78) = 1 - P(Z < 2.78)

                                                     = 1 - 0.9973 = 0.0027

The above probability is calculated by looking at the value of x = 2.78 in the z table which has an area of 0.9973.

Therefore, P(79 < < 81) = 0.9973 - 0.0027 = 0.9946.

(b) Probability that the sample mean diameter exceeds 81 when n = 36 is given by = P(> 81)

          P( > 81) = P( > ) = P(Z > 3.33) = 1 - P(Z < 3.33)

                                                       = 1 - 0.9996 = 0.0004

The above probability is calculated by looking at the value of x = 3.33 in the z table which has an area of 0.9996.

Hence, it is 0.04% likely is it that the sample mean diameter exceeds 81 when n = 36.

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