HW04

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University Of Chicago *

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22000

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Statistics

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Apr 3, 2024

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HW04 2023-10-25 library (lattice) library (tidyverse) ## ── Attaching core tidyverse packages ──────────────────────── tidyverse 2.0.0 ── ## ✔ dplyr 1.1.3 ✔ readr 2.1.4 ## ✔ forcats 1.0.0 ✔ stringr 1.5.0 ## ✔ ggplot2 3.4.3 ✔ tibble 3.2.1 ## ✔ lubridate 1.9.3 ✔ tidyr 1.3.0 ## ✔ purrr 1.0.2 ## ── Conflicts ────────────────────────────────────────── tidyverse_conflicts() ── ## ✖ dplyr::filter() masks stats::filter() ## ✖ dplyr::lag() masks stats::lag() ## ℹ Use the conflicted package (<http://conflicted.r-lib.org/>) to force all conflicts to become errors Question 1) (a) Payout 0 = 1 - 0.21264 - 0.04325 - 0.00306 ## [1] 0.74105 ( - 1 * 0.74105 ) + ( 0 * 0.21264 ) + ( 2 * 0.04325 ) + ( 119 * 0.00306 ) ## [1] -0.29041 Thus expected payout is $ -0.29041 (c) standard dev first we calculate variance (((( - 1+0.29041 ) ^ 2 ) * 0.74105 ) + ((( 0 + 0.29041 ) ^ 2 ) * 0.21264 ) + ((( 2 + 0.29041 ) ^ 2 ) * 0.04325 ) + ((( 119 + 0.29041 ) ^ 2 ) * 0.00306 )) ## [1] 44.16237 std dev sqrt ( 44.16237 ) ## [1] 6.645477

std dev = 6.645 (d) if played 100 times expected net profit 100 * - 0.29041 ## [1] -29.041 -$29.041 (e) Since independant, variance of 100 days = 100 * 44.16237 ## [1] 4416.237 std dev = sqrt ( 4416.237 ) ## [1] 66.45477 std dev = 66.454 Question 2 EX = 35 * 1 / 38 - 1 * 37 / 38 EX ## [1] -0.05263158 Expected value = -1/19 VX = ( 1 / 38 * ( 35 + 1 / 19 ) ^ 2 ) + ( 37 / 38 * ( - 1 + 1 / 19 ) ^ 2 ) VX ## [1] 33.20776 Variance of x = 33.21 EY = 17 * 2 / 38 - 1 * 36 / 38 EY ## [1] -0.05263158 Expected value of Y = -1/19 VX = ( 2 / 38 * ( 17 + 1 / 19 ) ^ 2 ) + ( 36 / 38 * ( - 1 + 1 / 19 ) ^ 2 ) VX ## [1] 16.15512 Variance of Y = 16.15 (c) New expected value = -0.105 ( 34 * ( 1 / 38 )) + ( 16 * 2 / 38 ) + ( - 2 * 35 / 38 )

## [1] -0.1052632 New Variance = 47.57 ( 1 / 38 * ( 34 + 2 / 19 ) ^ 2 ) + ( 2 / 38 * ( 16 + 2 / 19 ) ^ 2 ) + ( 35 / 38 * ( - 2 + 2 / 19 ) ^ 2 ) ## [1] 47.56787 (d) variance of T = 47.57 V X + V Y = 33.21 + 16. 15 = 49.36 49.36 is not equal to 47.57 They are not the same since X and Y are not independent. Question 3 Drug A - ( 40 * 0.01 ) + ( 60 * 0.04 ) + ( 68 * 0.05 ) + ( 70 * 0.8 ) + ( 72 * 0.05 ) + ( 80 * 0.04 ) + ( 100 * 0.01 ) ## [1] 70 Expected value for drug A = 70 Drug B - ( 40 * 0.4 ) + ( 60 * 0.05 ) + ( 68 * 0.04 ) + ( 70 * 0.02 ) + ( 72 * 0.04 ) + ( 80 * 0.05 ) + ( 100 * 0.4 ) ## [1] 70 Expected value for Drug B - 70 (b) Variance for Drug A = 26.4 ( 30 ^ 2 * ( 0.01 )) + ( 10 ^ 2 * ( 0.04 )) + ( 2 ^ 2 * ( 0.05 )) + ( 0 ^ 2 * ( 0.8 )) + ( 2 ^ 2 * ( 0.05 )) + ( 10 ^ 2 * ( 0.04 )) + ( 30 ^ 2 * ( 0.01 )) ## [1] 26.4 Variance for Drug B = 730.32 ( 30 ^ 2 * ( 0.4 )) + ( 10 ^ 2 * ( 0.05 )) + ( 2 ^ 2 * ( 0.04 )) + ( 0 ^ 2 * ( 0.02 )) + ( 2 ^ 2 * ( 0.04 )) + ( 10 ^ 2 * ( 0.05 )) + ( 30 ^ 2 * ( 0.4 )) ## [1] 730.32 (c) While both Drugs produce the same avergae heartbeat at 70, they would not have identical effects as Drug B varies significantly more than Drug A. This variety means results have are less consistent than Drug A.

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Question 4 ( 60-55 ) / 15.5 ## [1] 0.3225806 P(Z > 0.32) 1-P(Z < 0.32) Using z-score table , 0.3745 37.45% (b) to be above 40 , 40-mean/SD = 40-55 / 15.5 = -0.97 to be below 60 , 60-mean/SD = 60-55 / 15.5 = 0.32 P(-0.97 < Z < 0.32) = P(Z < 0.32) - P(-0.97 < Z) = 0.6244 - 0.1660 = 0.4595 45.95% (c) For P = 0.85 , z = 1.04 1.04 = (x - 55) / 15.5 x = 71.12 (d) For P = 0.2 , z = -0.84 -0.84 = (x - 55) / 15.5 x = 41.98 Question 5 (a) u = 10 , v = 9 , sd = 3 z = (15 - 10)/3 = 1.67 P(z<1.67) = 0.9522 95.22% chance (b) z = (12 -10)/3 = 0.67 P(z>0.67) = 1 - P(z<0.67) = 1- 0.7486 = 0.2514 25.14 % chance (c) For P = 0.05 , z= -1.645 -1.645 = (x - 10)/ 3 = 5.065 fastest 5% of repairs take 5.065 hours

Question 6 bdims = read.table ( "https://www.openintro.org/data/tab-delimited/bdims.txt" , header= TRUE ) fdims = subset (bdims, sex == 0 ) fhgtmean = mean (fdims $ hgt) fhgtsd = sd (fdims $ hgt) (b) Taller than 5’ 7 norm dist = 20.78% Taller according to data = 18.46 % 1 - pnorm ( q = 170.2 , mean = fhgtmean, sd = fhgtsd) ## [1] 0.2078056 with (fdims, sum (hgt > 170.2 ) / length (hgt)) ## [1] 0.1846154 (c) Norm dist 25% percentile = 160.45 According to data = 160 qnorm ( 0.25 , mean= fhgtmean, sd= fhgtsd) ## [1] 160.458 quantile (fdims $ hgt, prob= 0.25 ) ## 25% ## 160 library (lattice) library (tidyverse) hist (fdims $ age, breaks= 10 , freq= F, xlab= "Women's Age in year" ) fagemean = mean (fdims $ age) fagesd = sd (fdims $ age) x = seq ( 0 , 70 , length.out= 100 ); y = dnorm (x, fagemean, fagesd) lines (x, y)

(e) From data = 40.38% Normal Dist = 33.51% Since the actual data did not follow a normal distribution, I did not expect the two percentages to be equal with (fdims, sum (age < 25 ) / length (age)) ## [1] 0.4038462 pnorm ( 25 , mean = fagemean, sd = fagesd) ## [1] 0.3351456

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