Question

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

Answer 1

(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.

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
`\bar X` = sample mean diameter

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

Z =
`(\bar X-\mu)/((\sigma)/(√(n) ) )` ~ N(0,1)

where,
`\mu` = population mean = 80 GPA

`\sigma` = standard deviation = 1.8 GPA

n = sample size = 25

(a) Probability that sample mean lies between 79 and 81 is given by = P(79 <
`\bar X` < 81) = P(
`\bar X` < 81) - P(
`\bar X`
`\leq` 79)

P(
`\bar X` < 81) = P(
`(\bar X-\mu)/((\sigma)/(√(n) ) )` <
`(81-80)/((1.8)/(√(25) ) )` ) = P(Z < 2.78) = 0.9973

P(
`\bar X`
`\leq` 79) = P(
`(\bar X-\mu)/((\sigma)/(√(n) ) )`
`\leq`
`(79-80)/((1.8)/(√(25) ) )` ) = P(Z
`\leq` -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 <
`\bar X` < 81) = 0.9973 - 0.0027 = 0.9946.

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

P(
`\bar X` > 81) = P(
`(\bar X-\mu)/((\sigma)/(√(n) ) )` >
`(81-80)/((1.8)/(√(36) ) )` ) = 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.