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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

A binary communication channel transmits a sequence of 鈥渂its鈥� (\({\bf{0s}}{\rm{ }}{\bf{and}}{\rm{ }}{\bf{1s}}\)). Suppose that for any particular bit transmitted, there is a\({\bf{10}}\% \)chance of a transmission error (a zero becoming a one or a one becoming a zero). Assume that bit errors occur independently of one another.

a. Consider transmitting\(1000\)bits. What is the approximate probability that at most\(125\)transmission errors occur?

b. Suppose the same\(1000\)-bit message is sent two different times independently of one another. What is the approximate probability that the number of errors in the first transmission is within\(50\)of the number of errors in the second?

A stockroom currently has \({\rm{30}}\) components of a certain type, of which \({\rm{8}}\) were provided by supplier \({\rm{1,10}}\)by supplier \({\rm{2}}\) , and \({\rm{12}}\) by supplier \({\rm{3}}\). Six of these are to be randomly selected for a particular assembly. Let \({\rm{X = }}\) the number of supplier l's components selected, \({\rm{Y = }}\) the number of supplier \({\rm{2}}\) 's components selected, and \({\rm{p(x,y)}}\) denote the joint pmf of \({\rm{X}}\) and\({\rm{Y}}\).

a. What is \({\rm{p(3,2)}}\) ? (Hint: Each sample of size \({\rm{6}}\) is equally likely to be selected. Therefore, \({\rm{p(3,2) = }}\) (number of outcomes with \({\rm{X = 3}}\) and \({\rm{Y = 2)/(}}\) total number of outcomes). Now use the product rule for counting to obtain the numerator and denominator.)

b. Using the logic of part (a), obtain \({\rm{p(x,y}}\) ). (This can be thought of as a multivariate hypergeometric distribution-sampling without replacement from a finite population consisting of more than two categories.)

Answer

The National Health Statistics Reports dated Oct. \({\rm{22, 2008}}\), stated that for a sample size of \({\rm{277 18 - }}\)year-old American males, the sample mean waist circumference was \({\rm{86}}{\rm{.3cm}}\). A somewhat complicated method was used to estimate various population percentiles, resulting in the following values:

a. Is it plausible that the waist size distribution is at least approximately normal? Explain your reasoning. If your answer is no, conjecture the shape of the population distribution.

b. Suppose that the population mean waist size is \({\rm{85cm}}\)and that the population standard deviation is \({\rm{15cm}}\). How likely is it that a random sample of \({\rm{277}}\) individuals will result in a sample mean waist size of at least \({\rm{86}}{\rm{.3cm}}\)?

c. Referring back to (b), suppose now that the population mean waist size in \({\rm{82cm}}\).Now what is the (approximate) probability that the sample mean will be at least \({\rm{86}}{\rm{.3cm}}\)? In light of this calculation, do you think that \({\rm{82cm}}\)is a reasonable value for \({\rm{\mu }}\)?

Carry out a simulation experiment using a statistical computer package or other software to study the sampling distribution of \({\rm{\bar X}}\) when the population distribution is Weibull with \({\rm{\alpha = 2}}\) and\({\rm{\beta = 5}}\), as in Example\({\rm{5}}{\rm{.20}}\).[A1] Consider the four sample sizes, and\({\rm{30}}\), and in each case use \({\rm{1000}}\) replications. For which of these sample sizes does the \({\rm{\bar X}}\) sampling distribution appear to be approximately normal?

The mean weight of luggage checked by a randomly selected tourist-class passenger flying between two cities on a certain airline is\({\bf{40}}\)lb, and the standard deviation is\({\bf{10}}\)lb. The mean and standard deviation for a business class passenger is\({\bf{30}}\)lb and\({\bf{6}}\)lb, respectively.

a. If there are\({\bf{12}}\)business-class passengers and\({\bf{50}}\)tourist-class passengers on a particular flight, what is the expected value of total luggage weight and the standard deviation of total luggage weight?

b. If individual luggage weights are independent, normally distributed RVs, what is the probability that total luggage weight is at most\({\bf{2500}}\)lb?
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