
Essential Statistics (2nd Edition)
2nd Edition
ISBN: 9780134134406
Author: Robert Gould, Colleen N. Ryan, Rebecca Wong
Publisher: PEARSON
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Chapter 1, Problem 56CRE
To determine
Construct a two-way table that summarizes the data.
Compare the percentage of “Better” for the antivenom group and the placebo group.
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3. [15] The joint PDF of RVS X and Y is given by
fx.x(x,y) = {
x) = { c(x +
{
c(x+y³),
0,
0≤x≤ 1,0≤ y ≤1
otherwise
where c is a constant.
(a) Find the value of c.
(b) Find P(0 ≤ X ≤,
Need help please
7. [10] Suppose that Xi, i = 1,..., 5, are independent normal random variables, where
X1, X2 and X3 have the same distribution N(1, 2) and X4 and X5 have the same
distribution N(-1, 1). Let
(a) Find V(X5 - X3).
1
= √(x1 + x2) — — (Xx3 + x4 + X5).
(b) Find the distribution of Y.
(c) Find Cov(X2 - X1, Y).
-
Chapter 1 Solutions
Essential Statistics (2nd Edition)
Ch. 1 - Prob. 1SECh. 1 - The data in Table 1A were collected from one of...Ch. 1 - Prob. 3SECh. 1 - The data in Table 1A were collected from one of...Ch. 1 - The data in Table 1A were collected from one of...Ch. 1 - Prob. 6SECh. 1 - The data in Table 1A were collected from one of...Ch. 1 - The data in Table 1A were collected from one of...Ch. 1 - Prob. 9SECh. 1 - The data in Table 1A were collected from one of...
Ch. 1 - Prob. 11SECh. 1 - Students’ Ages The accompanying table gives ages...Ch. 1 - Prob. 13SECh. 1 - Prob. 14SECh. 1 - Older Siblings (Example 3) At a small four-year...Ch. 1 - Prob. 16SECh. 1 - Prob. 17SECh. 1 - Prob. 18SECh. 1 - Prob. 19SECh. 1 - Prob. 20SECh. 1 - Prob. 21SECh. 1 - Prob. 22SECh. 1 - Population Prediction The 2009 World Almanac and...Ch. 1 - Prob. 24SECh. 1 - Prob. 25SECh. 1 - Prob. 26SECh. 1 - Prob. 27SECh. 1 - Prob. 28SECh. 1 - Prob. 29SECh. 1 - Prob. 30SECh. 1 - Prob. 31SECh. 1 - Prob. 32SECh. 1 - Prob. 33SECh. 1 - Prob. 34SECh. 1 - Prob. 35SECh. 1 - Prob. 36SECh. 1 - Prob. 37SECh. 1 - Prob. 38SECh. 1 - Prob. 39SECh. 1 - Prob. 40SECh. 1 - Prob. 41SECh. 1 - Prob. 42SECh. 1 - Prob. 43SECh. 1 - Prob. 44SECh. 1 - Prob. 45SECh. 1 - Prob. 46SECh. 1 - Prob. 47SECh. 1 - Prob. 48SECh. 1 - Prob. 49SECh. 1 - Prob. 50SECh. 1 - Prob. 51SECh. 1 - Prob. 52SECh. 1 - Prob. 53CRECh. 1 - Prob. 54CRECh. 1 - Prob. 55CRECh. 1 - Prob. 56CRECh. 1 - Prob. 57CRECh. 1 - Prob. 58CRECh. 1 - Prob. 59CRECh. 1 - Prob. 60CRECh. 1 - Prob. 61CRECh. 1 - Prob. 62CRE
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- 1. [10] Suppose that X ~N(-2, 4). Let Y = 3X-1. (a) Find the distribution of Y. Show your work. (b) Find P(-8< Y < 15) by using the CDF, (2), of the standard normal distribu- tion. (c) Find the 0.05th right-tail percentage point (i.e., the 0.95th quantile) of the distri- bution of Y.arrow_forward6. [10] Let X, Y and Z be random variables. Suppose that E(X) = E(Y) = 1, E(Z) = 2, V(X) = 1, V(Y) = V(Z) = 4, Cov(X,Y) = -1, Cov(X, Z) = 0.5, and Cov(Y, Z) = -2. 2 (a) Find V(XY+2Z). (b) Find Cov(-x+2Y+Z, -Y-2Z).arrow_forward1. [10] Suppose that X ~N(-2, 4). Let Y = 3X-1. (a) Find the distribution of Y. Show your work. (b) Find P(-8< Y < 15) by using the CDF, (2), of the standard normal distribu- tion. (c) Find the 0.05th right-tail percentage point (i.e., the 0.95th quantile) of the distri- bution of Y.arrow_forward
- == 4. [10] Let X be a RV. Suppose that E[X(X-1)] = 3 and E(X) = 2. (a) Find E[(4-2X)²]. (b) Find V(-3x+1).arrow_forward2. [15] Let X and Y be two discrete RVs whose joint PMF is given by the following table: y Px,y(x, y) -1 1 3 0 0.1 0.04 0.02 I 2 0.08 0.2 0.06 4 0.06 0.14 0.30 (a) Find P(X ≥ 2, Y < 1). (b) Find P(X ≤Y - 1). (c) Find the marginal PMFs of X and Y. (d) Are X and Y independent? Explain (e) Find E(XY) and Cov(X, Y).arrow_forward32. Consider a normally distributed population with mean μ = 80 and standard deviation σ = 14. a. Construct the centerline and the upper and lower control limits for the chart if samples of size 5 are used. b. Repeat the analysis with samples of size 10. 2080 101 c. Discuss the effect of the sample size on the control limits.arrow_forward
- Consider the following hypothesis test. The following results are for two independent samples taken from the two populations. Sample 1 Sample 2 n 1 = 80 n 2 = 70 x 1 = 104 x 2 = 106 σ 1 = 8.4 σ 2 = 7.6 What is the value of the test statistic? If required enter negative values as negative numbers (to 2 decimals). What is the p-value (to 4 decimals)? Use z-table. With = .05, what is your hypothesis testing conclusion?arrow_forwardPeriodically, Merrill Lynch customers are asked to evaluate Merrill Lynch financial consultants and services (2000 Merrill Lynch Client Satisfaction Survey). Higher ratings on the client satisfaction survey indicate better service with 7 the maximum service rating. Independent samples of service ratings for two financial consultants are summarized here. Consultant A has 10 years of experience, whereas consultant B has 1 year of experience. Use = .05 and test to see whether the consultant with more experience has the higher population mean service rating. Consultant A Consultant B = 16 = 10 = 6.82 = 6.25 = .64 = .75 State the null and alternative hypotheses.H0: 1 - 2 Ha: 1 - 2 Compute the value of the test statistic (to 2 decimals). What is the p-value?The p-value is What is your conclusion?arrow_forwardA firm paid its first annual dividend yesterday in the amount of $.15 per share. The company plans to double the dividend in each of the next 3 years. Starting in Year 4, the firm plans to pay $1.50 per share indefinitely. What is one share of this stock worth today if the market rate of return on similar securities is 13.8 percent? Multiple Choice $11.79 $8.92 $10.77 $11.02 $10.26arrow_forward
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