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Chapter 4: Q47E (page 169)

The weight distribution of parcels sent in a certain manner is normal with mean value\({\rm{12lb}}\)and standard deviation\({\rm{3}}{\rm{.5lb}}\). The parcel service wishes to establish a weight value\({\rm{c}}\)beyond which there will be a surcharge. What value of\({\rm{c}}\)is such that\({\rm{99\% }}\)of all parcels are at least\({\rm{1lb}}\)under the surcharge weight?

Short Answer

Expert verified

The value of c is \({\rm{21}}{\rm{.155}}\).

Step by step solution

01

Introduction

A specific instance of the normal distribution is the standard normal distribution. In a typical normal distribution, the mean is equal to zero\(\left( {{\rm{\mu = 0}}} \right)\), and the standard deviation is equal to one\({\rm{1 }}\left( {{\rm{\sigma = 1}}} \right)\). The random variable with the standard normal distribution is designated by the letter z.

02

Explanation

Let the weight of the parcels be represented by a \({\rm{rv}}\)\({\rm{X}}\)then it's mean and standard deviation are given as:

\(\begin{array}{l}{\rm{\mu = 12lb}}\\{\rm{\sigma = 3}}{\rm{.5lb}}\end{array}\)

It is given that \({\rm{99\% }}\)of all the values are below the \({\rm{(c - 1)lb}}\)limit. Hence, we can say that the \({\rm{z}}\)-value corresponding to it is equal to\({{\rm{Z}}_{{\rm{0}}{\rm{.01}}}}\), i.e., it is \({\rm{9}}{{\rm{9}}^{{\rm{th }}}}\)percentile of the distribution.

\({{\rm{z}}_{{\rm{0}}{\rm{.01}}}}\)means that area to the left of \({{\rm{z}}_{{\rm{0}}{\rm{.01}}}}\)under standard normal distribution curve is \({\rm{0}}{\rm{.99}}\)

we can also say that:

\(\phi \left( {{z_{0.01}}} \right) = 0.99\)

Where \(\phi (z)\)is the \({\rm{cdf}}\) of standard normal distributed \({\rm{rv Z}}\).

We check Appendix Table A. 3 to see if \(\phi (z)\)is equal to \({\rm{0}}{\rm{.99}}\)for any\({\rm{z}}\). From there

\({{\rm{z}}_{{\rm{0}}{\rm{.01}}}}{\rm{\gg 2}}{\rm{.33}}\)Now we know that the value corresponding to \({\rm{z}}\)-score of \({{\rm{Z}}_{{\rm{0}}{\rm{.01}}}}\) is\(\left( {{\rm{c - 1}}} \right)\), Hence

\(\begin{array}{l}\frac{{{\rm{(c - 1) - 12}}}}{{{\rm{3}}{\rm{.5}}}}{\rm{ = }}{{\rm{Z}}_{{\rm{0}}{\rm{.01}}}}\\\frac{{{\rm{c - 13}}}}{{{\rm{3}}{\rm{.5}}}}{\rm{ = 2}}{\rm{.33}}\\{\rm{c = 13 + (3}}{\rm{.5)(2}}{\rm{.33)}}\\{\rm{c = 21}}{\rm{.155}}\end{array}\)

Therefore, the value of c is \({\rm{21}}{\rm{.155}}\).

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Most popular questions from this chapter

The article "A Probabilistic Model of Fracture in Concrete and Size Effects on Fracture Toughness" (Magazine of Concrete Res., \({\rm{1996: 311 - 320}}\)) gives arguments for why fracture toughness in concrete specimens should have a Weibull distribution and presents several histograms of data that appear well fit by superimposed Weibull curves. Consider the following sample of size \({\rm{n = 18}}\) observations on toughness for high strength concrete (consistent with one of the histograms); values of \({{\rm{p}}_{\rm{i}}}{\rm{ = (i - }}{\rm{.5)/18}}\) are also given.

\(\begin{array}{*{20}{c}}{{\rm{ Observation }}}&{{\rm{.47}}}&{{\rm{.58}}}&{{\rm{.65}}}&{{\rm{.69}}}&{{\rm{.72}}}&{{\rm{.74}}}\\{{{\rm{p}}_{\rm{i}}}}&{{\rm{.0278}}}&{{\rm{.0833}}}&{{\rm{.1389}}}&{{\rm{.1944}}}&{{\rm{.2500}}}&{{\rm{.3056}}}\\{{\rm{ Observation }}}&{{\rm{.77}}}&{{\rm{.79}}}&{{\rm{.80}}}&{{\rm{.81}}}&{{\rm{.82}}}&{{\rm{.84}}}\\{{{\rm{p}}_{\rm{i}}}}&{{\rm{.3611}}}&{{\rm{.4167}}}&{{\rm{.4722}}}&{{\rm{.5278}}}&{{\rm{.5833}}}&{{\rm{.6389}}}\\{{\rm{ Observation }}}&{{\rm{.86}}}&{{\rm{.89}}}&{{\rm{.91}}}&{{\rm{.95}}}&{{\rm{1}}{\rm{.01}}}&{{\rm{1}}{\rm{.04}}}\\{{{\rm{p}}_{\rm{i}}}}&{{\rm{.6944}}}&{{\rm{.7500}}}&{{\rm{.8056}}}&{{\rm{.8611}}}&{{\rm{.9167}}}&{{\rm{.9722}}}\end{array}\)

Construct a Weibull probability plot and comment.

A college professor never finishes his lecture before the end of the hour and always finishes his lectures within \({\rm{2}}\) min after the hour. Let \({\rm{X = }}\)the time that elapses between the end of the hour and the end of the lecture and suppose the pdf of \({\rm{X}}\) is

\({\rm{f(x) = \{ }}\begin{array}{*{20}{c}}{{\rm{k}}{{\rm{x}}^2}}&{{\rm{0}} \le {\rm{x}} \le {\rm{2}}}\\{\rm{0}}&{{\rm{otherwise}}}\end{array}\)

a. Find the value of \({\rm{k}}\) and draw the corresponding density curve. (Hint: Total area under the graph of \({\rm{f(x)}}\) is \({\rm{1}}\).)

b. What is the probability that the lecture ends within \({\rm{1}}\) min of the end of the hour?

c. What is the probability that the lecture continues beyond the hour for between \({\rm{60}}\) and \({\rm{90}}\) sec?

d. What is the probability that the lecture continues for at least \({\rm{90}}\) sec beyond the end of the hour?

a. Show that if X has a normal distribution with parameters \({\rm{\mu }}\) and\({\rm{\sigma }}\), then \({\rm{Y = aX + b}}\) (a linear function of X ) also has a normal distribution. What are the parameters of the distribution of Y (i.e., E(Y) and V(Y)) ? (Hint: Write the cdf of\({\rm{Y,P(Y}} \le {\rm{y)}}\), as an integral involving the pdf of X, and then differentiate with respect to y to get the pdf of Y.)

b. If, when measured in\(^{\rm{^\circ }}{\rm{C}}\), temperature is normally distributed with mean 115 and standard deviation 2 , what can be said about the distribution of temperature measured in\(^{\rm{^\circ }}{\rm{F}}\)?

The mode of a continuous distribution is the value \({{\rm{x}}^{\rm{*}}}\) that maximizes\({\rm{f(x)}}\).

a. What is the mode of a normal distribution with parameters \({\rm{\mu }}\) and \({\rm{\sigma }}\) ?

b. Does the uniform distribution with parameters \({\rm{A}}\) and \({\rm{B}}\) have a single mode? Why or why not?

c. What is the mode of an exponential distribution with parameter\({\rm{\lambda }}\)? (Draw a picture.)

d. If \({\rm{X}}\) has a gamma distribution with parameters \({\rm{\alpha }}\) and\({\rm{\beta }}\), and\({\rm{\alpha > 1}}\), find the mode. (Hint: \({\rm{ln(f(x))}}\)will be maximized if \({\rm{f(x)}}\) is, and it may be simpler to take the derivative of\({\rm{ln(f(x))}}\).)

e. What is the mode of a chi-squared distribution having \({\rm{v}}\) degrees of freedom?

Consider babies born in the 鈥渘ormal鈥� range of\(37 - 43\)weeks gestational age. Extensive data support the assumption that for such babies born in the United States, birth weight is normally distributed with a mean of\(3432 g\)and a standard deviation of\(482\)g. (The article 鈥淎re Babies Normal?鈥� analyzed data from a particular year; for a sensible choice of class intervals, a histogram did not look at all normal, but after further investigations it was determined that this was due to some hospitals measuring weight in grams and others measuring to the nearest ounce and then converting to grams. A modified choice of class intervals that allowed for this gave a histogram that was well described by a normal distribution.) a. What is the probability that the birth weight of a randomly selected baby of this type exceeds\(4000 g\)? Is between\(3000 and 4000 g\)? b. What is the probability that the birth weight of a randomly selected baby of this type is either less than\(2000 g\)or greater than\(5000 g\)? c. What is the probability that the birth weight of a randomly selected baby of this type exceeds\(7\)lb? d. How would you characterize the most extreme\(.1\% \)of all birth weights? e. If X is a random variable with a normal distribution and a is a numerical constant, then Y = X also has a normal distribution. Use this to determine the distribution of birth weight expressed in pounds (shape, mean, and standard deviation), and then recalculate the probability from part (c). How does this compare to your previous answer?
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