Beginning Statistics, 2nd Edition
2nd Edition
ISBN: 9781932628678
Author: Carolyn Warren; Kimberly Denley; Emily Atchley
Publisher: Hawkes Learning Systems
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Question
Chapter 4.4, Problem 47E
To determine
To Choose:
The greatest number of different five-letter arrangements from the following words.
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Students have asked these similar questions
9. The concentration function of a random variable X is defined as
Qx(h) = sup P(x ≤ X ≤x+h), h>0.
x
(a) Show that Qx+b (h) = Qx(h).
(b) Is it true that Qx(ah) =aQx(h)?
(c) Show that, if X and Y are independent random variables, then
Qx+y (h) min{Qx(h). Qy (h)).
To put the concept in perspective, if X1, X2, X, are independent, identically
distributed random variables, and S₁ = Z=1Xk, then there exists an absolute
constant, A, such that
A
Qs, (h) ≤
√n
Some references: [79, 80, 162, 222], and [204], Sect. 1.5.
29
Suppose that a mound-shaped data set has a
must mean of 10 and standard deviation of 2.
a. About what percentage of the data should
lie between 6 and 12?
b. About what percentage of the data should
lie between 4 and 6?
c. About what percentage of the data should
lie below 4?
91002 175/1
3
2,3,
ample
and
rical
t?
the
28 Suppose that a mound-shaped data set has a
mean of 10 and standard deviation of 2.
a. About what percentage of the data should
lie between 8 and 12?
b. About what percentage of the data should
lie above 10?
c. About what percentage of the data should
lie above 12?
Chapter 4 Solutions
Beginning Statistics, 2nd Edition
Ch. 4.1 - Prob. 1ECh. 4.1 - Prob. 2ECh. 4.1 - Prob. 3ECh. 4.1 - Prob. 4ECh. 4.1 - Prob. 5ECh. 4.1 - Prob. 6ECh. 4.1 - Prob. 7ECh. 4.1 - Prob. 8ECh. 4.1 - Prob. 9ECh. 4.1 - Prob. 10E
Ch. 4.1 - Prob. 11ECh. 4.1 - Prob. 12ECh. 4.1 - Prob. 13ECh. 4.1 - Prob. 14ECh. 4.1 - Prob. 15ECh. 4.1 - Prob. 16ECh. 4.1 - Prob. 17ECh. 4.1 - Prob. 18ECh. 4.1 - Prob. 19ECh. 4.1 - Prob. 20ECh. 4.1 - Prob. 21ECh. 4.1 - Prob. 22ECh. 4.1 - Prob. 23ECh. 4.1 - Prob. 24ECh. 4.1 - Prob. 25ECh. 4.1 - Prob. 26ECh. 4.2 - Prob. 1ECh. 4.2 - Prob. 2ECh. 4.2 - Prob. 3ECh. 4.2 - Prob. 4ECh. 4.2 - Prob. 5ECh. 4.2 - Prob. 6ECh. 4.2 - Prob. 7ECh. 4.2 - Prob. 8ECh. 4.2 - Prob. 9ECh. 4.2 - Prob. 10ECh. 4.2 - Prob. 11ECh. 4.2 - Prob. 12ECh. 4.2 - Prob. 13ECh. 4.2 - Prob. 14ECh. 4.2 - Prob. 15ECh. 4.2 - Prob. 16ECh. 4.2 - Prob. 17ECh. 4.2 - Prob. 18ECh. 4.2 - Prob. 19ECh. 4.2 - Prob. 20ECh. 4.2 - Prob. 21ECh. 4.2 - Prob. 22ECh. 4.2 - Prob. 23ECh. 4.2 - Prob. 24ECh. 4.2 - Prob. 25ECh. 4.2 - Prob. 26ECh. 4.2 - Prob. 27ECh. 4.2 - Prob. 28ECh. 4.2 - Prob. 29ECh. 4.2 - Prob. 30ECh. 4.2 - Prob. 31ECh. 4.3 - Prob. 1ECh. 4.3 - Prob. 2ECh. 4.3 - Prob. 3ECh. 4.3 - Prob. 4ECh. 4.3 - Prob. 5ECh. 4.3 - Prob. 6ECh. 4.3 - Prob. 7ECh. 4.3 - Prob. 8ECh. 4.3 - Prob. 9ECh. 4.3 - Prob. 10ECh. 4.3 - Prob. 11ECh. 4.3 - Prob. 12ECh. 4.3 - Prob. 13ECh. 4.3 - Prob. 14ECh. 4.3 - Prob. 15ECh. 4.3 - Prob. 16ECh. 4.3 - Prob. 17ECh. 4.3 - Prob. 18ECh. 4.3 - Prob. 19ECh. 4.3 - Prob. 20ECh. 4.3 - Prob. 21ECh. 4.3 - Prob. 22ECh. 4.3 - Prob. 23ECh. 4.3 - Prob. 24ECh. 4.3 - Prob. 25ECh. 4.3 - Prob. 26ECh. 4.3 - Prob. 27ECh. 4.3 - Prob. 28ECh. 4.3 - Prob. 29ECh. 4.3 - Prob. 30ECh. 4.3 - Prob. 31ECh. 4.4 - Prob. 1ECh. 4.4 - Prob. 2ECh. 4.4 - Prob. 3ECh. 4.4 - Prob. 4ECh. 4.4 - Prob. 5ECh. 4.4 - Prob. 6ECh. 4.4 - Prob. 7ECh. 4.4 - Prob. 8ECh. 4.4 - Prob. 9ECh. 4.4 - Prob. 10ECh. 4.4 - Prob. 11ECh. 4.4 - Prob. 12ECh. 4.4 - Prob. 13ECh. 4.4 - Prob. 14ECh. 4.4 - Prob. 15ECh. 4.4 - Prob. 16ECh. 4.4 - Prob. 17ECh. 4.4 - Prob. 18ECh. 4.4 - Prob. 19ECh. 4.4 - Prob. 20ECh. 4.4 - Prob. 21ECh. 4.4 - Prob. 22ECh. 4.4 - Prob. 23ECh. 4.4 - Prob. 24ECh. 4.4 - Prob. 25ECh. 4.4 - Prob. 26ECh. 4.4 - Prob. 27ECh. 4.4 - Prob. 28ECh. 4.4 - Prob. 29ECh. 4.4 - Prob. 30ECh. 4.4 - Prob. 31ECh. 4.4 - Prob. 32ECh. 4.4 - Prob. 33ECh. 4.4 - Prob. 34ECh. 4.4 - Prob. 35ECh. 4.4 - Prob. 36ECh. 4.4 - Prob. 37ECh. 4.4 - Prob. 38ECh. 4.4 - Prob. 39ECh. 4.4 - Prob. 40ECh. 4.4 - Prob. 41ECh. 4.4 - Prob. 42ECh. 4.4 - Prob. 43ECh. 4.4 - Prob. 44ECh. 4.4 - Prob. 45ECh. 4.4 - Prob. 46ECh. 4.4 - Prob. 47ECh. 4.4 - Prob. 48ECh. 4.4 - Prob. 49ECh. 4.4 - Prob. 50ECh. 4.4 - Prob. 51ECh. 4.4 - Prob. 52ECh. 4.5 - Prob. 1ECh. 4.5 - Prob. 2ECh. 4.5 - Prob. 3ECh. 4.5 - Prob. 4ECh. 4.5 - Prob. 5ECh. 4.5 - Prob. 6ECh. 4.5 - Prob. 7ECh. 4.5 - Prob. 8ECh. 4.5 - Prob. 9ECh. 4.5 - Prob. 10ECh. 4.5 - Prob. 11ECh. 4.5 - Prob. 12ECh. 4.5 - Prob. 13ECh. 4.5 - Prob. 14ECh. 4.CR - Prob. 1CRCh. 4.CR - Prob. 2CRCh. 4.CR - Prob. 3CRCh. 4.CR - Prob. 4CRCh. 4.CR - Prob. 5CRCh. 4.CR - Prob. 6CRCh. 4.CR - Prob. 7CRCh. 4.CR - Prob. 8CRCh. 4.CR - Prob. 9CRCh. 4.CR - Prob. 10CRCh. 4.CR - Prob. 11CRCh. 4.CR - Prob. 12CRCh. 4.CR - Prob. 13CRCh. 4.CR - Prob. 14CRCh. 4.CR - Prob. 15CRCh. 4.CR - Prob. 16CRCh. 4.CR - Prob. 17CRCh. 4.CR - Prob. 18CRCh. 4.CR - Prob. 19CRCh. 4.CR - Prob. 20CRCh. 4.CR - Prob. 21CRCh. 4.CR - Prob. 22CRCh. 4.CR - Prob. 23CRCh. 4.CR - Prob. 24CRCh. 4.CR - Prob. 25CR
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Similar questions
- 27 Suppose that you have a data set of 1, 2, 2, 3, 3, 3, 4, 4, 5, and you assume that this sample represents a population. The mean is 3 and g the standard deviation is 1.225.10 a. Explain why you can apply the empirical rule to this data set. b. Where would "most of the values" in the population fall, based on this data set?arrow_forward30 Explain how you can use the empirical rule to find out whether a data set is mound- shaped, using only the values of the data themselves (no histogram available).arrow_forward5. Let X be a positive random variable with finite variance, and let A = (0, 1). Prove that P(X AEX) 2 (1-A)² (EX)² EX2arrow_forward
- 6. Let, for p = (0, 1), and xe R. X be a random variable defined as follows: P(X=-x) = P(X = x)=p. P(X=0)= 1-2p. Show that there is equality in Chebyshev's inequality for X. This means that Chebyshev's inequality, in spite of being rather crude, cannot be improved without additional assumptions.arrow_forward4. Prove that, for any random variable X, the minimum of EIX-al is attained for a = med (X).arrow_forward8. Recall, from Sect. 2.16.4, the likelihood ratio statistic, Ln, which was defined as a product of independent, identically distributed random variables with mean 1 (under the so-called null hypothesis), and the, sometimes more convenient, log-likelihood, log L, which was a sum of independent, identically distributed random variables, which, however, do not have mean log 1 = 0. (a) Verify that the last claim is correct, by proving the more general statement, namely that, if Y is a non-negative random variable with finite mean, then E(log Y) log(EY). (b) Prove that, in fact, there is strict inequality: E(log Y) < log(EY), unless Y is degenerate. (c) Review the proof of Jensen's inequality, Theorem 5.1. Generalize with a glimpse on (b).arrow_forward
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- The Honolulu Advertiser stated that in Honolulu there was an average of 659 burglaries per 400,000 households in a given year. In the Kohola Drive neighborhood there are 321 homes. Let r be the number of homes that will be burglarized in a year. Use the formula for Poisson distribution. What is the value of p, the probability of success, to four decimal places?arrow_forwardThe college hiking club is having a fundraiser to buy new equipment for fall and winter outings. The club is selling Chinese fortune cookies at a price of $2 per cookie. Each cookie contains a piece of paper with a different number written on it. A random drawing will determine which number is the winner of a dinner for two at a local Chinese restaurant. The dinner is valued at $32. Since fortune cookies are donated to the club, we can ignore the cost of the cookies. The club sold 718 cookies before the drawing. Lisa bought 13 cookies. Lisa's expected earnings can be found by multiplying the value of the dinner by the probability that she will win. What are Lisa's expected earnings? Round your answer to the nearest cent.arrow_forwardWhat was the age distribution of nurses in Great Britain at the time of Florence Nightingale? Thanks to Florence Nightingale and the British census of 1851, we have the following information (based on data from the classic text Notes on Nursing, by Florence Nightingale). Note: In 1851 there were 25,466 nurses in Great Britain. Furthermore, Nightingale made a strict distinction between nurses and domestic servants. Use a histogram and graph the probability distribution. Using the graph of the probability distribution determine the probability that a British nurse selected at random in 1851 would be 40 years of age or older. Round your answer to nearest thousandth. Age range (yr) 20–29 30–39 40–49 50–59 60–69 70–79 80+ Midpoint (x) 24.5 34.5 44.5 54.5 64.5 74.5 84.5 Percent of nurses 5.7% 9.7% 19.5% 29.2% 25.0% 9.1% 1.8%arrow_forward
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