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Chapter 5: Q47E (page 237)

Refer to Exercise \({\rm{46}}\). Suppose the distribution is normal (the cited article makes that assumption and even includes the corresponding normal density curve).

a. Calculate \({\rm{P}}\)(\(69\text{拢}\bar{X}\text{拢}71\)) when \({\rm{n = 16}}\).

b. How likely is it that the sample mean diameter exceeds \({\rm{71}}\) when \({\rm{n = 25}}\)?

Short Answer

Expert verified
  1. \(\text{P(69拢 x虅 拢71)=0}\text{.9876=98}\text{.76% }\)
  2. \({\rm{P(\bar X > 71) = 0}}{\rm{.0009 = 0}}{\rm{.09\% }}\)

Step by step solution

01

Definition of young’s modulus

Young's Modulus (also known as Elastic Modulus or Tensile Modulus) is a mechanical property of linear elastic solids such as rods, wires, and other similar objects. Other quantities, such as Bulk modulus and shear modulus, can be used to determine a material's elastic properties, but Young's Modulus is the most widely utilised.

02

Calculating (a) when \({\rm{n = 16}}\)

Given:

\(\begin{aligned}&{{\rm{\mu = 70GPa}}}\\&{{\rm{\sigma = 1}}{\rm{.6GPa}}}\\&{{\rm{n = 16}}}\end{aligned}\)

The sample mean's sampling distribution has a mean of \({\rm{\mu }}\) and a standard deviation of \(\frac{{\rm{\sigma }}}{{\sqrt {\rm{n}} }}\)

The \({\rm{z}}\)-value is the sample mean divided by the standard deviation of the population mean:

\(\begin{aligned}&{{\rm{z = }}\frac{{{\rm{\bar x - \mu }}}}{{{\rm{\sigma /}}\sqrt {\rm{n}} }}{\rm{ = }}\frac{{{\rm{69 - 70}}}}{{{\rm{1}}{\rm{.6/}}\sqrt {{\rm{16}}} }}{\rm{\gg - 2}}{\rm{.50}}}\\&{{\rm{z = }}\frac{{{\rm{\bar x - \mu }}}}{{{\rm{\sigma /}}\sqrt {\rm{n}} }}{\rm{ = }}\frac{{{\rm{71 - 70}}}}{{{\rm{1}}{\rm{.6/}}\sqrt {{\rm{16}}} }}{\rm{\gg 2}}{\rm{.50}}}\end{aligned}\)

Using the normal probability table in the appendix, calculate the corresponding probability (which contains the probability to the left of \({\rm{z}}\)-scores).

\(\begin{aligned}\rm P(69拢\bar X拢71) &= P( - 2{\rm{.50 < Z < 2}{\rm{.50) = P(Z < 2}}{\rm{.50) - P(Z < - 2}}{\rm{.50)}}}\\ \rm &= 0{\rm{.9938 - 0}}{\rm{.0062 = 0}}{\rm{.9876 = 98}}{\rm{.76\% }}\end{aligned}\)

03

Calculating the sample mean diameter exceeds \({\rm{71}}\) when \({\rm{n = 25}}\)

Given:

\({\rm{\mu = 70GPa}}\)

\({\rm{\sigma = 1}}{\rm{.6GPa}}\)

\({\rm{n = 25}}\)

The sample mean's sampling distribution has a mean of \({\rm{\mu }}\) and a standard deviation of \(\frac{{\rm{\sigma }}}{{\sqrt {\rm{n}} }}\)

The \({\rm{z}}\)-value is the sample mean divided by the standard deviation of the population mean:

\({\rm{z = }}\frac{{{\rm{\bar x - \mu }}}}{{{\rm{\sigma /}}\sqrt {\rm{n}} }}{\rm{ = }}\frac{{{\rm{71 - 70}}}}{{{\rm{1}}{\rm{.6/}}\sqrt {{\rm{25}}} }}{\rm{\gg 3}}{\rm{.13}}\)

Using the normal probability table in the appendix, calculate the corresponding probability (which contains the probability to the left of \({\rm{z}}\)-scores).

\(\begin{aligned}\rm P(\bar X > 71) &= P(Z < 3{\rm{.13) = 1 - P(Z < 3}}{\rm{.13)}}\\\rm &= 1 - 0{\rm{.9991 = 0}}{\rm{.0009 = 0}}{\rm{.09\% }}\end{aligned}\)

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Most popular questions from this chapter

We have seen that if\(E\left( {{X_1}} \right) = E\left( {{X_2}} \right) = . . . = E\left( {{X_n}} \right) = \mu \), then\(E\left( {{X_1} + . . . + {X_n}} \right) = n\mu \). In some applications, the number of\({X_i} 's\)under consideration is not a fixed number n but instead is a rv N. For example, let N\(5\)be the number of components that are brought into a repair shop on a particular day, and let Xi denote the repair shop time for the\({i^{th}}\)component. Then the total repair time is\({X_1} + {X_2} + . . . + {X_n}\), the sum of a random number of random variables. When N is independent of the\({X_i} 's\), it can be shown that

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