7.8.1 Suppose the random variables ...are i.i.d. (independent and identically distributed), and consider the sum .... Find the mean and variance of Sn, and write and sketch the p.d.
f. of the approximate normal distribution. Suppose that n =16, and that ...takes the value 0 with probability 0.5 and the value 1 with probability 0.5.
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7.8.2 Suppose the random variables ...are i.i.d. (independent and identically distributed), and consider the sum .... Find the mean and variance of Sn, and write and sketch the p.d.
f. of the approximate normal distribution. Suppose that n =50, and that ...takes the value 0 with probability 0.25, the value 1 with probability 0.5, and the value 2 with probability 0.25.
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7.8.3 Suppose the random variables ...are i.i.d. and consider the average .... Find the mean and variance of ..., and write and sketch the p.d.
f. of the approximate normal distribution. Suppose that n =16, and that Xi takes the value 0 with probability 0.5 and the value 1 with probability 0.5 (as in Exercise 1).
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7.8.4 Suppose the random variables ...are i.i.d. and consider the average .... Find the mean and variance of ..., and write and sketch the p.d.
f. of the approximate normal distribution. Suppose that n =50, and that ... takes
the value 0 with probability 0.25, the value 1 with probability 0.5,
and the value 2 with probability 0.25( as in Exercise 2).
Get solution
7.8.5
Recall
the data describing the probabilities of the outcomes of four
experiments counting the number of mutants in a bacterial culture
(Section 6.6, Exercises 1–4). ... Suppose that each experiment
is repeated (independently) and the total number of mutants is counted.
Experiment a is
repeated 20 times. Write an integral that estimates the probability
that there are between 30 and 40 mutants, and shade the corresponding
area on a sketch of the approximate normal p.d.
f. Take a guess at the
probability based on your sketch. (We found the expectation in Section
6.7, Exercise 1 and the variance in Section 6.9, Exercise 1.)
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7.8.6 Recall
the data describing the probabilities of the outcomes of four
experiments counting the number of mutants in a bacterial culture
(Section 6.6, Exercises 1–4). ... Experiment b is
repeated 80 times. Write an integral that estimates the probability
that there are between 30 and 40 mutants, and shade the corresponding
area on a sketch of the approximate normal p.d.
f. Take a guess at the
probability based on your sketch. (We found the expectation in Section
6.7, Exercise 2 and the variance in Section 6.9, Exercise 2.)
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7.8.7 Recall
the data describing the probabilities of the outcomes of four
experiments counting the number of mutants in a bacterial culture
(Section 6.6, Exercises 1–4). ... Experiment c is
repeated 100 times. Write an integral that estimates the probability
that there is an average of less than 1.8 mutants per experiment and
shade the corresponding area on a sketch of the approximate normal
p.d.
f. Take a guess at the probability based on your sketch. (We found
the expectation in Section 6.7, Exercise 3 and the variance in Section
6.9, Exercise 3.)
Get solution
7.8.9 Suppose that X and Y are independent normally distributed random variables with X ≈ N(5.0, 16.0) and Y ≈ N(10.0, 9.0). Find and sketch the p.d.
f. of 3X.
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7.8.9 Suppose that X and Y are independent normally distributed random variables with X ≈ N(5.0, 16.0) and Y ≈ N(10.0, 9.0). Find and sketch the p.d.
f. of 3X.
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7.8.10 Suppose that X and Y are independent normally distributed random variables with X ≈ N(5.0, 16.0) and Y ≈ N(10.0, 9.0). Find and sketch the p.d.
f. of X + Y .
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7.8.11 Suppose that X and Y are independent normally distributed random variables with X ≈ N(5.0, 16.0) and Y ≈ N(10.0, 9.0). Find and sketch the p.d.
f. of the sum of nine independent samples from X .
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7.8.12 Suppose that X and Y are independent normally distributed random variables with X ≈ N(5.0, 16.0) and Y ≈ N(10.0, 9.0). Find and sketch the p.d.
f. of the mean of nine independent samples from Y .
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7.8.13 Show the following facts about the normal distribution. The normal p.d.
f. with μ=0 and ... =1 takes on its maximum at x =0.
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7.8.14 Show the following facts about the normal distribution. The normal p.d.
f. takes on its maximum at x =μ for any values of μ and σ .
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7.8.15 Show the following facts about the normal distribution. The normal p.d.
f. with μ=0 and ... =1 has points of inflection at x =−1 and x =1.
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7.8.16 Show the following facts about the normal distribution. The normal p.d.
f. has points of inflection at x =μ + σ and x =μ − σ for any values of μ and σ.
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7.8.17 ... Use the law of total probability to show that ...
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7.8.18 ... Evaluate the integral to find ...
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7.8.19 ... Use the same trick to find ...
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7.8.20 ... Can you guess the pattern? Why does the answer look so much like the Poisson distribution?
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7.8.21 The
Central Limit Theorem does not work if random variables are not
independent. The simplest case to compute is where they are perfectly
correlated with each other. In particular, suppose the random variables ...are all equal, and consider the sum ...Find the mean and variance of ..., and sketch its probability distribution. Suppose that n =16 and that ...takes the value 0 with probability 0.5 and the value 1 with probability 0.5. Compare with the results in Exercise 1.
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7.8.23 Based on the probabilities in Figure 7.8.1b, we can find the probability that a plant gains 5 cm in height from ten genes, Pr(H =5). Find the two ways to add up 0’s, 1’s, and 2.5’s to get 5 cm.
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7.8.23 Based on the probabilities in Figure 7.8.1b, we can find the probability that a plant gains 5 cm in height from ten genes, Pr(H =5). Find the two ways to add up 0’s, 1’s, and 2.5’s to get 5 cm.
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7.8.24 Based on the probabilities in Figure 7.8.1b, we can find the probability that a plant gains 5 cm in height from ten genes, Pr(H =5). Find the number of ways each could occur (use binomial coefficients).
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7.8.25 Based on the probabilities in Figure 7.8.1b, we can find the probability that a plant gains 5 cm in height from ten genes, Pr(H =5). Find the probability associated with each way.
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7.8.26 Based on the probabilities in Figure 7.8.1b, we can find the probability that a plant gains 5 cm in height from ten genes, Pr(H =5). Add them up to find the total probability.
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7.8.27 Based on the probabilities in Figure 7.8.1b, we can find the probability that a plant gains 5 cm in height from ten genes, Pr(H =5). Find the normal distribution approximating added height, and find the value of the normal p.d.
f. at 5.0 cm.
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7.8.28 Based on the probabilities in Figure 7.8.1b, we can find the probability that a plant gains 5 cm in height from ten genes, Pr(H =5). Use
the result of the previous problem to approximate the probability that
the height is in the interval between 4.5 and 5.5 cm. Compare with the
result in Exercise 26.
Get solution
7.8.29 Suppose
that scientists develop a new model of human IQ that includes three
independent factors: genes of large effect, genes of small effect, and
environmental effects. All genes are assumed to be dominant. There are
ten smart genes of large effect, each of which adds 2.5 IQ points. There
are 20 smart genes of small effect, each of which adds 0.6 IQ points.
There are 30 environmental factors, with favorable ones adding 0.9 IQ
points. People have a baseline IQ of 80 if they have no favorable
effects. Finally, suppose that the probability of getting each smart
gene is 0.75, and the probability of getting each favorable
environmental effect is 0.5. Find the normal approximation for IQ based only on genes of large effect.
Get solution
7.8.30 Suppose
that scientists develop a new model of human IQ that includes three
independent factors: genes of large effect, genes of small effect, and
environmental effects. All genes are assumed to be dominant. There are
ten smart genes of large effect, each of which adds 2.5 IQ points. There
are 20 smart genes of small effect, each of which adds 0.6 IQ points.
There are 30 environmental factors, with favorable ones adding 0.9 IQ
points. People have a baseline IQ of 80 if they have no favorable
effects. Finally, suppose that the probability of getting each smart
gene is 0.75, and the probability of getting each favorable
environmental effect is 0.5. Find the normal approximation for IQ based only on genes of small effect.
Get solution
7.8.31 Suppose
that scientists develop a new model of human IQ that includes three
independent factors: genes of large effect, genes of small effect, and
environmental effects. All genes are assumed to be dominant. There are
ten smart genes of large effect, each of which adds 2.5 IQ points. There
are 20 smart genes of small effect, each of which adds 0.6 IQ points.
There are 30 environmental factors, with favorable ones adding 0.9 IQ
points. People have a baseline IQ of 80 if they have no favorable
effects. Finally, suppose that the probability of getting each smart
gene is 0.75, and the probability of getting each favorable
environmental effect is 0.5. Find the normal approximation for IQ based only on environmental effects.
Get solution
7.8.32 Suppose
that scientists develop a new model of human IQ that includes three
independent factors: genes of large effect, genes of small effect, and
environmental effects. All genes are assumed to be dominant. There are
ten smart genes of large effect, each of which adds 2.5 IQ points. There
are 20 smart genes of small effect, each of which adds 0.6 IQ points.
There are 30 environmental factors, with favorable ones adding 0.9 IQ
points. People have a baseline IQ of 80 if they have no favorable
effects. Finally, suppose that the probability of getting each smart
gene is 0.75, and the probability of getting each favorable
environmental effect is 0.5. Find the normal approximation for
IQ with both genetic and environmental effects. What is the maximum
possible IQ with the model?
Get solution
7.8.33 Suppose
immigration and emigration change the sizes of four populations with
the following probabilities (Section 6.7, Exercises 31–34). Find the
p.d.
f. of the normal approximation for the average number of immigrants,
sketch a graph, and shade and estimate the area corresponding to an
increase in the population. ... Suppose immigrants arrive into and emigrate from population a for 20 yr.
Get solution
7.8.34 Suppose
immigration and emigration change the sizes of four populations with
the following probabilities (Section 6.7, Exercises 31–34). Find the
p.d.
f. of the normal approximation for the average number of immigrants,
sketch a graph, and shade and estimate the area corresponding to an
increase in the population. ... Suppose immigrants arrive into and emigrate from population b for 20 yr.
Get solution
7.8.35 Suppose
immigration and emigration change the sizes of four populations with
the following probabilities (Section 6.7, Exercises 31–34). Find the
p.d.
f. of the normal approximation for the average number of
immigrants,
sketch a graph, and shade and estimate the area corresponding to an
increase in the population. ... Suppose immigrants arrive into
and emigrate from population c for 10 yr. How accurate do you think the
normal approximation is?
Get solution
7.8.36 Suppose
immigration and emigration change the sizes of four populations with
the following probabilities (Section 6.7, Exercises 31–34). Find the
p.d.
f. of the normal approximation for the average number of
immigrants,
sketch a graph, and shade and estimate the area corresponding to an
increase in the population. ... Suppose immigrants arrive into
and emigrate from population d for 10 yr. How accurate do you think the
normal approximation is?
Get solution
7.8.37 Although the Central Limit Theorem applies to the sums of independent and identically distributed random variables, we can use logarithms to analyze products of
independent and identically distributed random variables. Consider
populations growing for the given number of years with the given
distribution for the random variable R giving per capita production. Find the normal distribution that approximates the logarithm of the population size ... assuming that ... =100. ... =4 with probability 0.5, ... =0.25 with probability 0.5 (as in Section 6.8, Exercise 37). Find the approximate normal distribution for ln(...), the log of the population size after 50 time steps.
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7.8.38 Although the Central Limit Theorem applies to the sums of independent and identically distributed random variables, we can use logarithms to analyze products of
independent and identically distributed random variables. Consider
populations growing for the given number of years with the given
distribution for the random variable R giving per capita production. Find the normal distribution that approximates the logarithm of the population size ... assuming that ... =100. ... =4 with probability 0.25, ... =0.25 with probability 0.75 (as in Section 6.8, Exercise 38). Find the approximate normal distribution for ln(...).
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7.8.39 As
in Exercise, the Central Limit Theorem for sums can be used to
approximate the logarithm of a product even when the random variables
multiplied together are continuous random variables. Find the normal
distribution that approximates the logarithm of the population size ... assuming that ... =1. You will need the indefinite integrals ... to evaluate the expectation and variance of ln(R).
Use the rule of thumb that most populations end up within two standard
deviations from the mean to give a range of probable population sizes. Let R be a random variable giving the per capita production in a population with p.d.
f. g(x)=5.0 for 1.0 ≤ x ≤1.2 (the values used in Example 6.1.2). Are the simulations in Example 6.1.2 within this range?
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7.8.40 As
in Exercise, the Central Limit Theorem for sums can be used to
approximate the logarithm of a product even when the random variables
multiplied together are continuous random variables. Find the normal
distribution that approximates the logarithm of the population size ... assuming that ... =1. You will need the indefinite integrals ... to evaluate the expectation and variance of ln(R).
Use the rule of thumb that most populations end up within two standard
deviations from the mean to give a range of probable population sizes. Let R be a random variable giving the per capita production in a population with p.d.
f. g(x)=1.25 for 0.7 ≤ x ≤1.5 (the values used in Example 6.1.3). Are the simulations in Example 6.1.3 within this range?
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7.8.41 Two other values often used to describe distributions are the skewness and the kurtosis. Suppose a random variable has p.d.
f. f (x) for −∞< x <∞ and mean μ. The skewness ...
describes symmetry, and has a formula like the variance: ...
Instead of taking the difference from the mean and squaring it, we take
the difference from the mean and cube it. This is sometimes called the third moment around the mean. Symmetric distributions have a skewness of 0. The kurtosis ...is based on the fourth moment around the mean. However, it includes a correction factor involving the variance ... ... The correction was added to make the kurtosis of a normal distribution equal to 0.
a. Find the skewness and kurtosis of a normal distribution with mean 0 and variance 1.
b. Find the skewness and kurtosis of a normal distribution with mean 1 and variance 2. c.
Find the skewness and kurtosis of an exponential distribution with mean
1.0. (Remember that the random variable in this case must be positive.)
The skewness is positive because the distribution is stretched out to
the right.
d. Find the skewness and kurtosis of a uniform distribution with p.d.
f. f (x)=1 for 0 ≤ x ≤1. The kurtosis is positive because the distribution has “broad shoulders.”
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7.8.42 Program your computer to choose random numbers from an exponential distribution with mean 1. a.
Pick 100 such numbers, and put them into 15 bins (the number between 0
and 0.2, the number between 0.2 and 0.4, etc). Plot a histogram of the
number in each bin.
b. Pick 100 pairs of random numbers and take the average of each pair. Put these values into bins and plot a histogram.
c. Pick 100 sets of ten random numbers and take the average of each set. Put these values into bins and plot a histogram. d.
Pick 100 sets of 30 random numbers and take the average of each set.
Put these values into bins and plot a histogram. Do the results look
more and more like a normal distribution?
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