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Chapter 5: Q53E (page 237)

Rockwell hardness of pins of a certain type is known to have a mean value of 50 and a standard deviation of \({\rm{1}}{\rm{.2}}{\rm{.}}\)

a. If the distribution is normal, what is the probability that the sample mean hardness for a random sample of \({\rm{9}}\) pins is at least \({\rm{51}}\)?

b. Without assuming population normality, what is the (approximate) probability that the sample mean hardness for a random sample of \({\rm{40 }}\) pins is at least \({\rm{51}}\)?

Short Answer

Expert verified

a.${\text{P(\bar X51) = 0}}{\text{.0062}}$

b.${\text{P(\bar X51) = 0}}$

Step by step solution

01

Definition of probability

the proportion of the total number of conceivable outcomes to the number of options in an exhaustive collection of equally likely outcomes that cause a given occurrence.

02

Determining the probability that the sample mean hardness for a random sample of \({\rm{9}}\) pins is at least \({\rm{51}}\)

Let \({\rm{X}}\)be a random variable with a \({\rm{\mu = 50}}\)mean and a \({\rm{\sigma = 1}}{\rm{.2}}\) standard deviation. The sample average's (mean) \({\rm{\bar X}}\)is \({{\rm{\mu }}_{{\rm{\bar X}}}}{\rm{ = \mu = 50}}\), while the sample average's (mean) standard deviation is

\({{\rm{\sigma }}_{{\rm{\bar X}}}}{\rm{ = }}\frac{{\rm{1}}}{{\sqrt {\rm{n}} }}{\rm{ \times \sigma = }}\frac{{\rm{1}}}{{\sqrt {\rm{9}} }}{\rm{ \times 1}}{\rm{.2 = 0}}{\rm{.4,}}\;\;\;{\rm{n = 9\;pins}}{\rm{.\;}}\)

The following statement is correct:

\[{{\text{P(\bar X51) = P}}\left( {\frac{{{\text{\bar X - }}{{\text{\mu }}_{{\text{\bar X}}}}}}{{{{\text{\sigma }}_{{\text{\bar X}}}}}}{\text{}}\frac{{{\text{51 - 50}}}}{{{\text{0}}{\text{.4}}}}} \right){\text{ = P(Z2}}{\text{.5)}}}\]

\[{{\text{ = 1 - P(Z < 2}}{\text{.5)}}\mathop {\text{ = }}\limits^{{\text{(1)}}} {\text{1 - 0}}{\text{.9938}}}\]

\[{{\text{ = 0}}{\text{.0062,}}}\]

(1): from the appendix's normal probability table. Software can also be used to calculate the likelihood.

03

 Determining the probability that the sample mean hardness for a random sample of \({\rm{40}}\) pins is at least \({\rm{51}}\)

Let \({\rm{X}}\)be a random variable with a \({\rm{\mu = 50}}\)mean and a \({\rm{\sigma = 1}}{\rm{.2}}\) standard deviation. The sample average's (mean) \({\rm{\bar X}}\)is \({{\rm{\mu }}_{{\rm{\bar X}}}}{\rm{ = \mu = 50}}\), while the sample average's (mean) standard deviation is

\({{\rm{\sigma }}_{{\rm{\bar X}}}}{\rm{ = }}\frac{{\rm{1}}}{{\sqrt {\rm{n}} }}{\rm{ \times \sigma = }}\frac{{\rm{1}}}{{\sqrt {{\rm{40}}} }}{\rm{ \times 1}}{\rm{.2 = 0}}{\rm{.19,}}\;\;\;{\rm{n = 40\;pins}}{\rm{.\;}}\)

The following statement is correct:

\[{{\text{P(\bar X51) = P}}\left( {\frac{{{\text{\bar X - }}{{\text{\mu }}_{{\text{\bar X}}}}}}{{{{\text{\sigma }}_{{\text{\bar X}}}}}}{\text{}}\frac{{{\text{51 - 50}}}}{{{\text{0}}{\text{.19}}}}} \right){\text{ = P(Z5}}{\text{.27)}}}\]

\[{{\text{ = 1 - P(Z < 5}}{\text{.27)}}\mathop {\text{ = }}\limits^{{\text{(1)}}} {\text{1 - 1}}}\]

\[{{\text{ = 0}}}\]

(1): from the appendix's normal probability table. Software can also be used to calculate the likelihood.

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Most popular questions from this chapter

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If two loads are applied to a cantilever beam as shown in the accompanying drawing, the bending moment at \({\rm{0}}\) due to the loads is \({{\rm{a}}_{\rm{1}}}{{\rm{X}}_{\rm{1}}}{\rm{ + }}{{\rm{a}}_{\rm{2}}}{{\rm{X}}_{\rm{2}}}{\rm{ }}\)

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