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Chapter 5: Q73E (page 243)

Suppose the expected tensile strength of type-A steel is \({\rm{105ksi}}\)and the standard deviation of tensile strength is \({\rm{8ksi}}\). For type-B steel, suppose the expected tensile strength and standard deviation of tensile strength are \({\rm{100ksi}}\)and \({\rm{6ksi}}\), respectively. Let \({\rm{\bar X = }}\)the sample average tensile strength of a random sample of \({\rm{40}}\) type-A specimens, and let \({\rm{\bar Y = }}\)the sample average tensile strength of a random sample of \({\rm{35}}\)type-B specimens.

a. What is the approximate distribution of \({\rm{\bar X ? of \bar Y?}}\)

b. What is the approximate distribution of \({\rm{\bar X - \bar Y}}\)? Justify your answer.

c. Calculate (approximately) \(P( - 1拢\bar X - \bar Y拢1)\)

d. Calculate. If you actually observed , would you doubt that \({{\rm{\mu }}_{\rm{1}}}{\rm{ - }}{{\rm{\mu }}_{\rm{2}}}{\rm{ = 5?}}\)

Short Answer

Expert verified

(a): \({\rm{ \bar X}}\)Approximately normal, with a mean of \({\rm{105}}\)and a standard deviation of \({\rm{1}}{\rm{.2649}}{\rm{.}}\)

\({\rm{\bar Y}}\): Almost typical, with a mean of \({\rm{100}}\) and a standard deviation of $1.0142.

(b) \({\rm{\bar X - \bar Y}}\): Close to normal, with a mean of \({\rm{5}}\) and a standard deviation of \({\rm{1}}{\rm{.6213}}\)

(c) \({\rm{P( - 1拢\bar X - \bar Y拢1) = 0}}{\rm{.0068 = 0}}{\rm{.68\% }}\)

(d) We're not convinced that \({{\rm{\mu }}_{\rm{1}}}{\rm{ - }}{{\rm{\mu }} _ {\rm{2}}}{\rm{ = 5}}\)

Step by step solution

01

Definition of standard deviation

The square root of the variance is the standard deviation of a random variable, sample, statistical population, data collection, or probability distribution. It is less resilient in practice than the average absolute deviation, but it is algebraically easier.

02

Determining the approximate distribution of \({\rm{\bar X ,\bar Y}}\)

Given:

\(\begin{array}{*{20}{c}} {{\mu _X} = 105}\\ {{\sigma _X} = 8}\\ {{\mu _Y} = 100}\\ {{\sigma _Y} = 6}\\ {{n_X} = 40}\\ {{n_Y} = 35} \end{array}\)

The central limit theorem states that if the sample size is big (\({\rm{30}}\) or more), the sample mean \({\rm{\bar x}}\)sampling distribution is approximately normal.

The central limit theorem tells us that the sampling distribution of the sample mean \({\rm{\bar x}}\)is nearly normal because the sample size of \({\rm{40}}\)is at least \({\rm{30}}\)

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

\(\begin{array}{*{20}{c}}{{{\rm{\mu }}_{{\rm{\bar x}}}}{\rm{ = }}{{\rm{\mu }}_{\rm{X}}}{\rm{ = 105}}}\\{{{\rm{\sigma }}_{{\rm{\bar x}}}}{\rm{ = }}\frac{{{{\rm{\sigma }}_{\rm{X}}}}}{{\sqrt {\rm{n}} }}{\rm{ = }}\frac{{\rm{8}}}{{\sqrt {{\rm{40}}} }}{\rm{ = }}\frac{{{\rm{2}}\sqrt {{\rm{10}}} }}{{\rm{5}}}{\rm{\gg 1}}{\rm{.2649}}}\end{array}\)

The central limit theorem states that if the sample size is big (\({\rm{30}}\) or more), the sample mean \({\rm{\bar y}}\)sampling distribution is essentially normal.

The central limit theorem tells us that the sampling distribution of the sample mean \({\rm{\bar y}}\)is nearly normal because the sample size of \({\rm{35}}\) is at least \({\rm{30}}\)

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

\(\begin{array}{*{20}{c}}{{{\rm{\mu }}_{{\rm{\bar y}}}}{\rm{ = }}{{\rm{\mu }}_{\rm{Y}}}{\rm{ = 100}}}\\{{{\rm{\sigma }}_{{\rm{\bar y}}}}{\rm{ = }}\frac{{{{\rm{\sigma }}_{\rm{Y}}}}}{{\sqrt {\rm{n}} }}{\rm{ = }}\frac{{\rm{6}}}{{\sqrt {{\rm{35}}} }}{\rm{ = }}\frac{{{\rm{6}}\sqrt {{\rm{35}}} }}{{{\rm{35}}}}{\rm{\gg 1}}{\rm{.0142}}}\end{array}\)

03

 Determining the approximate distribution of \({\rm{\bar X - \bar Y}}\)

For the linear combination \({\rm{W = a}}{{\rm{X}}_{\rm{1}}}{\rm{ + b}}{{\rm{X}}_{\rm{2}}}\),the mean, variance, and standard deviation have the following properties:

\(\begin{array}{*{20}{c}}{{{\rm{\mu }}_{\rm{W}}}{\rm{ = a}}{{\rm{\mu }}_{\rm{1}}}{\rm{ + b}}{{\rm{\mu }}_{\rm{2}}}}\\{{\rm{\sigma }}_{\rm{W}}^{\rm{2}}{\rm{ = }}{{\rm{a}}^{\rm{2}}}{\rm{\sigma }}_{\rm{1}}^{\rm{2}}{\rm{ + }}{{\rm{b}}^{\rm{2}}}{\rm{\sigma }}_{\rm{2}}^{\rm{2}}\left( {{\rm{\;If\;}}{{\rm{X}}_{\rm{ - }}}{\rm{1\;and\;}}{{\rm{X}}_{\rm{ - }}}{\rm{2\;are independent\;}}} \right)}\\{{{\rm{\sigma }}_{\rm{W}}}{\rm{ = }}\sqrt {{{\rm{a}}^{\rm{2}}}{\rm{\sigma }}_{\rm{1}}^{\rm{2}}{\rm{ + }}{{\rm{b}}^{\rm{2}}}{\rm{\sigma }}_{\rm{2}}^{\rm{2}}} \left( {{\rm{\;If\;}}{{\rm{X}}_{\rm{ - }}}{\rm{1\;and\;}}{{\rm{X}}_{\rm{ - }}}{\rm{2}}} \right.{\rm{\;are independent)\;}}}\end{array}\)

Suppose, \({\rm{\bar Xand \bar Y}}\) are independent

\({{\rm{\mu }}_{{\rm{\bar x - \bar y}}}}{\rm{ = }}{{\rm{\mu }}_{{\rm{\bar x}}}}{\rm{ - }}{{\rm{\mu }}_{{\rm{\bar y}}}}{\rm{ = 105 - 100 = 5}}\)

\({{\rm{\sigma }}_{{\rm{\bar x - \bar y}}}}{\rm{ = }}\sqrt {{{\rm{\sigma }}_{{\rm{\bar x}}}}{\rm{ + ( - 1}}{{\rm{)}}^{\rm{2}}}{{\rm{\sigma }}_{{\rm{\bar y}}}}} {\rm{ = }}\sqrt {{\rm{1}}{\rm{.264}}{{\rm{9}}^{\rm{2}}}{\rm{ + 1}}{\rm{.014}}{{\rm{2}}^{\rm{2}}}} {\rm{\gg 1}}{\rm{.6213}}\)

Because \({\rm{\bar Xand \bar Y}}\) are approximately normally distributed, \({\rm{\bar X - \bar Y}}\)is about normally distributed (and because we assume that they are independent).

04

Calculating (approximately) \(P( - 1£\bar X - \bar Y£1)\)

The standardized score is calculated by dividing the value \({\rm{x}}\)by the mean and then by the standard deviation.

\(\begin{array}{*{20}{c}}{{\rm{z = }}\frac{{{\rm{x - \mu }}}}{{\rm{\sigma }}}{\rm{ = }}\frac{{{\rm{1 - 5}}}}{{{\rm{1}}{\rm{.6213}}}}{\rm{\gg - 2}}{\rm{.47}}}\\{{\rm{z = }}\frac{{{\rm{x - \mu }}}}{{\rm{\sigma }}}{\rm{ = }}\frac{{{\rm{ - 1 - 5}}}}{{{\rm{1}}{\rm{.6213}}}}{\rm{\gg - 3}}{\rm{.70}}}\end{array}\)

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

\(\begin{array}{*{20}{c}}{{\rm{P( - 1拢\bar X - \bar Y拢1) = P( - 3}}{\rm{.70 < Z < - 2}}{\rm{.47) = P(Z < - 2}}{\rm{.47) - P(Z < - 3}}{\rm{.70)}}}\\{{\rm{\gg 0}}{\rm{.0068 - 0 = 0}}{\rm{.0068 = 0}}{\rm{.68\% }}}\end{array}\)

05

Calculating

The standardized score is the value \({\rm{x}}\)divided by the standard deviation after being reduced by the mean.

\({\rm{z = }}\frac{{{\rm{x - \mu }}}}{{\rm{\sigma }}}{\rm{ = }}\frac{{{\rm{10 - 5}}}}{{{\rm{1}}{\rm{.6213}}}}{\rm{\gg 3}}{\rm{.08}}\)

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

We would be skeptical of \({{\rm{\mu }}_{\rm{1}}}{\rm{ - }}{{\rm{\mu }}_{\rm{2}}}{\rm{ = 5}}\)because the probability is so small. if was obtained.

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Most popular questions from this chapter

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?

Show that if\({\rm{X}}\)and\({\rm{Y}}\)are independent rv's, then\({\rm{E(XY) = E(X) \times E(Y)}}\). Then apply this in Exercise\({\rm{25}}{\rm{.}}\)[A1] (Hint: Consider the continuous case with\({\rm{f(x,y) = }}\)\({{\rm{f}}_{\rm{X}}}{\rm{(x) \times }}{{\rm{f}}_{\rm{Y}}}{\rm{(y)}}\).)

Consider a small ferry that can accommodate cars and buses. The toll for cars is\({\rm{\$ 3}}\), and the toll for buses is\({\rm{\$ 10}}\). Let \({\rm{X}}\) and \({\rm{Y}}\) denote the number of cars and buses, respectively, carried on a single trip. Suppose the joint distribution of \({\rm{X}}\) and\({\rm{Y}}\). Compute the expected revenue from a single trip.

Let X1, X2, and X3 represent the times necessary to perform three successive repair tasks at a certain service facility. Suppose they are independent, normal rv鈥檚 with expected values \({\mu _1}, {\mu _2}, and {\mu _3}\)and variances \(\sigma _1^2 , \sigma _2^2, and \sigma _3^2 \), respectively. a. If \(\mu = {\mu _2} = {\mu _3} = 60\)and\(\sigma _1^2 = \sigma _2^2 = \sigma _3^2 = 15\), calculate \(P\left( {{T_0} \le 200} \right)\)and\(P\left( {150 \le {T_0} \le 200} \right)\)? b. Using the \(\mu 's and \sigma 's\)given in part (a), calculate both \(P\left( {55 \le X} \right)\)and \(P\left( {58 \le X \le 62} \right)\).c. Using the \(\mu 's and \sigma 's\)given in part (a), calculate and interpret\(P\left( { - 10 \le {X_1} - .5{X_2} - .5{X_3} \le 5} \right)\). d. If\({\mu _1} = 40, {\mu _1} = 50, {\mu _1} = 60,\),\( \sigma _1^2 = 10, \sigma _2^2 = 12, and \sigma _3^2 = 14\) calculate \(P\left( {{X_1} + {X_2} + {X_3} \le 160} \right)\)and also \(P\left( {{X_1} + {X_2} \ge 2{X_3}} \right).\)

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 \({{\rm{a}}_{\rm{1}}}{{\rm{X}}_{\rm{1}}}{\rm{ + }}{{\rm{a}}_{\rm{2}}}{{\rm{X}}_{\rm{2}}}{\rm{ }}\)

a. Suppose that \({{\rm{X}}_{\rm{1}}}{\rm{ and }}{{\rm{X}}_{\rm{2}}}{\rm{ }}\)are independent rv鈥檚 with means \({\rm{2 and 4kip}}\), respectively, and standard deviations \({\rm{.5}}\) and \({\rm{1}}{\rm{.0kip}}\), respectively. If \({{\rm{a}}_{\rm{1}}}{\rm{ = 5ft and }}{{\rm{a}}_{\rm{2}}}{\rm{ = 10ft }}\), what is the expected bending moment and what is the standard deviation of the bending moment?

b. If \({{\rm{X}}_{\rm{1}}}{\rm{ and }}{{\rm{X}}_{\rm{2}}}{\rm{ }}\)are normally distributed, what is the probability that the bending moment will exceed 75 kip-ft?

c. Suppose the positions of the two loads are random variables. Denoting them by \({{\rm{A}}_{\rm{1}}}{\rm{ and }}{{\rm{A}}_{\rm{2}}}{\rm{ }}\), assume that these variables have means of \({\rm{5 and 10ft }}\), respectively, that each has a standard deviation of \({\rm{.5}}\), and that all are independent of one another. What is the expected moment now?

d. For the situation of part (c), what is the variance of the bending moment?

e. If the situation is as described in part (a) except that \({\rm{Corr}}\left( {{{\rm{X}}_{\rm{1}}}{\rm{,}}{{\rm{X}}_{\rm{2}}}} \right){\rm{ = 0}}{\rm{.5}}\) (so that the two loads are not independent), what is the variance of the bending moment?
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