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Chapter 5: Q50E (page 237)

Let \({\rm{X}}\)denote the courtship time for a randomly selected female鈥搈ale pair of mating scorpion flies (time from the beginning of interaction until mating). Suppose the mean value of\({\rm{X}}\) is \({\rm{120}}\) min and the standard deviation of \({\rm{X}}\) is \({\rm{110}}\) min (suggested by data in the article 鈥淪hould I Stay or Should I Go? Condition- and Status-Dependent Courtship Decisions in the Scorpion Fly Panorpa Cognate鈥� (Animal Behavior, \({\rm{2009: 491 - 497}}\))). transmitted, there is a \({\rm{ 10\% }}\)chance of a transmission error (a \({\rm{0}}\)becoming a \({\rm{1}}\) or a \({\rm{1}}\)becoming a \({\rm{0}}\)). Assume that bit errors occur independently of one another.

a. Is it plausible that \({\rm{X}}\) is normally distributed?

b. For a random sample of \({\rm{50}}\) such pairs, what is the (approximate) probability that the sample mean courtship time is between \({\rm{100}}\) min and \({\rm{125}}\) min?

c. For a random sample of \({\rm{50}}\) such pairs, what is the (approximate) probability that the total courtship time exceeds \({\rm{150}}\) hrs.?

d. Could the probability requested in (b) be calculated from the given information if the sample size were \({\rm{15}}\) rather than \({\rm{50}}\)? Explain.

Short Answer

Expert verified

a. Not possible.

b. \({\rm{P(100 < \bar X < 125) = 0}}{\rm{.5270 = 52}}{\rm{.70\% }}\)

c. Probability is close to zero.

d. Because the sample size is fewer than \({\rm{30}}\), the answer is no.

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 that \({\rm{X}}\) is normally distributed        

Given:

\(\begin{array}{*{20}{c}}{{\rm{\mu = 120min}}}\\{{\rm{\sigma = 110min}}}\end{array}\)

For a randomly selected female-male pair of mating scorpion flies, \({\rm{X = }}\)courting time.

Because a negative time makes no sense in this scenario, the shortest conceivable time is \({\rm{0}}\) minutes.

The duration might be as long as possible, implying that the distribution is probably skewed to the right (or positively skewed).

Because a normal distribution can take on negative values and is not skewed, \({\rm{X}}\)cannot be considered a normal distribution

03

Determining probability that the sample mean courtship time is between \({\rm{100}}\) min and \({\rm{125}}\) min

Given:

\(\begin{array}{*{20}{c}}{{\rm{\mu = 120min}}}\\{{\rm{\sigma = 110min}}}\end{array}\)

\({\rm{n = 50}}\)

The central limit theorem states that if the sample size is big (more than \({\rm{30}}\)), 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 is greater than \({\rm{30}}\).

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:

\(\begin{array}{*{20}{c}}{{\rm{z = }}\frac{{{\rm{\bar x - \mu }}}}{{{\rm{\sigma /}}\sqrt {\rm{n}} }}{\rm{ = }}\frac{{{\rm{100 - 120}}}}{{{\rm{110/}}\sqrt {{\rm{50}}} }}{\rm{\gg - 1}}{\rm{.29}}}\\{{\rm{z = }}\frac{{{\rm{\bar x - \mu }}}}{{{\rm{\sigma /}}\sqrt {\rm{n}} }}{\rm{ = }}\frac{{{\rm{125 - 120}}}}{{{\rm{110/}}\sqrt {{\rm{50}}} }}{\rm{\gg 0}}{\rm{.32}}}\end{array}\)

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}P(100 拢 \bar X 拢125) &= P( - 1.29 < Z < 0.32) = P(Z < 0.32) - P(Z < - 1.29) \\& = 0.6255 - 0.0985 = 0.5270 = 52.70\% \end{aligned}

04

Determining the probability that the total courtship time exceeds \({\rm{150}}\) hr

Given:

\(\begin{array}{*{20}{c}}{{\rm{\mu = 120min}}}\\{{\rm{\sigma = 110min}}}\end{array}\)

\({\rm{n = 50}}\)

The total duration is \({\rm{150}}\)hours, or \({\rm{150 \times 60 = 9000}}\)minutes. The sample mean is calculated by dividing the total time by the sample size:

\({\rm{\bar x = }}\frac{{{\rm{9000}}}}{{{\rm{50}}}}{\rm{ = 180}}\)

The central limit theorem states that if the sample size is big (more than \({\rm{30}}\)), 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 is greater than \({\rm{30}}\).

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:

\({\rm{z = }}\frac{{{\rm{\bar x - \mu }}}}{{{\rm{\sigma /}}\sqrt {\rm{n}} }}{\rm{ = }}\frac{{{\rm{180 - 120}}}}{{{\rm{110/}}\sqrt {{\rm{50}}} }}{\rm{\gg 3}}{\rm{.86}}\)

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

\(P\left( {{a^{\circ } }{X_i} > 150} \right) = P(Z > 3.86) = 1 - P(Z < 3.86)\gg 1 - 1 = 0\)

05

calculating the sample size were \({\rm{15 }}\)rather than\({\rm{50 }}\)

Given:

\(\begin{array}{*{20}{c}}{{\rm{\mu = 120min}}}\\{{\rm{\sigma = 110min}}}\\{{\rm{n = 50}}}\\{{\rm{n = 15}}}\end{array}\)

The central limit theorem states that if the sample size is big (more than \({\rm{30}}\)), 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 is greater than \({\rm{30}}\).

We can't utilise the central limit theorem because the sample size is smaller than 30. It is not possible to compute the requested probability since the distribution of \({\rm{X}}\)is not nearly normal (by part (a)) and we do not know the distribution of \({\rm{X}}\)

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Most popular questions from this chapter

A health-food store stocks two different brands of a certain type of grain. Let \(X = \)the amount (lb) of brand A on hand and \(Y = \)the amount of brand B on hand. Suppose the joint pdf of X and Y is

\(f(x,y) = \left\{ {\begin{array}{*{20}{c}}{kxy\;\;\;\;\;\;\;x \ge 0,\;y \ge 0,\;20 \le x + y \le 30}\\{0\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;otherwise}\end{array}} \right\}\)

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a. What is the joint pmf of \({\rm{X}}\) and\({\rm{Y}}\)?

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\({\rm{P(X + Y = m)}}\mathop {\rm{ = }}\limits^{{\rm{(1)}}} {\sum\limits_{{\rm{k = 0}}}^{\rm{m}} {\left( {\begin{array}{*{20}{c}}{\rm{m}}\\{\rm{k}}\end{array}} \right){{\rm{a}}^{\rm{k}}}{{\rm{b}}^{{\rm{m - k}}}}{\rm{ = }}\left( {{\rm{a + b}}} \right)} ^{\rm{m}}}\)

Refer back to Example, Two cars with six-cylinder engines and three with four-cylinder engines are to be driven over a \(300\)-mile course. Let \({X_1}, . . . {X_5}\)denote the resulting fuel efficiencies (mpg). Consider the linear combination

\(Y = \left( {{X_1} + {X_2}} \right)/2 - \left( {{X_3} + {X_4} + {X_5}} \right)/3\)

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c. Why is the probability in (b) much larger than the probability in (a)?

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