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a.
To find: The number of outcomes in the
To find: The number of outcomes in the
To find: The number of outcomes in the event a child does not have the gene, given that the child has a normal IQ.
To find: The number of outcomes in the sample space of event a child does not have the gene, given that the child has a normal IQ.
b.
To divide: The number of outcomes in the event ‘a child does not have the gene’ by the number of outcomes in the sample space of event ‘a child does not have the gene’.
To divide: The number of outcomes in the event ‘a child does not have the gene, given that the child has a normal IQ’ with the number of outcomes in the sample space of event ‘a child does not have the gene, given that the child has a normal IQ’.
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Chapter 3 Solutions
Elementary Statistics Books a la carte Plus NEW MyLab Statistics with Pearson eText - Access Card Package (6th Edition)
- Examine the Variables: Carefully review and note the names of all variables in the dataset. Examples of these variables include: Mileage (mpg) Number of Cylinders (cyl) Displacement (disp) Horsepower (hp) Research: Google to understand these variables. Statistical Analysis: Select mpg variable, and perform the following statistical tests. Once you are done with these tests using mpg variable, repeat the same with hp Mean Median First Quartile (Q1) Second Quartile (Q2) Third Quartile (Q3) Fourth Quartile (Q4) 10th Percentile 70th Percentile Skewness Kurtosis Document Your Results: In RStudio: Before running each statistical test, provide a heading in the format shown at the bottom. “# Mean of mileage – Your name’s command” In Microsoft Word: Once you've completed all tests, take a screenshot of your results in RStudio and paste it into a Microsoft Word document. Make sure that snapshots are very clear. You will need multiple snapshots. Also transfer these results to the…arrow_forward2 (VaR and ES) Suppose X1 are independent. Prove that ~ Unif[-0.5, 0.5] and X2 VaRa (X1X2) < VaRa(X1) + VaRa (X2). ~ Unif[-0.5, 0.5]arrow_forward8 (Correlation and Diversification) Assume we have two stocks, A and B, show that a particular combination of the two stocks produce a risk-free portfolio when the correlation between the return of A and B is -1.arrow_forward
- 9 (Portfolio allocation) Suppose R₁ and R2 are returns of 2 assets and with expected return and variance respectively r₁ and 72 and variance-covariance σ2, 0%½ and σ12. Find −∞ ≤ w ≤ ∞ such that the portfolio wR₁ + (1 - w) R₂ has the smallest risk.arrow_forward7 (Multivariate random variable) Suppose X, €1, €2, €3 are IID N(0, 1) and Y2 Y₁ = 0.2 0.8X + €1, Y₂ = 0.3 +0.7X+ €2, Y3 = 0.2 + 0.9X + €3. = (In models like this, X is called the common factors of Y₁, Y₂, Y3.) Y = (Y1, Y2, Y3). (a) Find E(Y) and cov(Y). (b) What can you observe from cov(Y). Writearrow_forward1 (VaR and ES) Suppose X ~ f(x) with 1+x, if 0> x > −1 f(x) = 1−x if 1 x > 0 Find VaRo.05 (X) and ES0.05 (X).arrow_forward
- Joy is making Christmas gifts. She has 6 1/12 feet of yarn and will need 4 1/4 to complete our project. How much yarn will she have left over compute this solution in two different ways arrow_forwardSolve for X. Explain each step. 2^2x • 2^-4=8arrow_forwardOne hundred people were surveyed, and one question pertained to their educational background. The results of this question and their genders are given in the following table. Female (F) Male (F′) Total College degree (D) 30 20 50 No college degree (D′) 30 20 50 Total 60 40 100 If a person is selected at random from those surveyed, find the probability of each of the following events.1. The person is female or has a college degree. Answer: equation editor Equation Editor 2. The person is male or does not have a college degree. Answer: equation editor Equation Editor 3. The person is female or does not have a college degree.arrow_forward
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