Assume that random guesses are made for 3 multiplechoice questions on a test with 5 choices for each question, so that there are n = 3 trials, each with probability of success (correct) given by p = 0.20. Find the probability of no correct answers 3 Click on the icon to view the binomial probability table. The probability of no correct answers is (Round to three decimal places as needed.) .70 80 90 X 10 30 40 05 .20 002 0+ 640 490 360 250 160 090 .040 010 980 810 020 480 500 480 420 320 180 095 020 180 320 420 .640 .810 980 250 360 490 902 0+ 010 040 090 160 001 0+ -ON 3 027 008 729 .512 343 216 125 064 970 0+ 375 288 189 .096 027 007 029 135 .243 384 441 432 .135 .029 .288 432 441 .384 243 .375 0+ 007 .027 .096 .189 970 125 343 .729 857 216 .512 o+ D+ .001 .008 027 .064 OWN 0+ 0+ 815 656 .410 240 .130 062 026 008 .002 961 .026 0+ 076 .004 0+ 171 292 .410 412 .346 250 154 039 265 154 .049 .014 001 014 .049 .154 265 .346 .375 346 001 412 410 292 171 039 076 .154 250 346 OWN 0+ 004 026 .656 815 961 0+ .002 .008 .026 062 130 240 .410 0+ o+ .078 002 04 031 010 951 774 590 328 .168 0+ o+ 259 156 077 028 006 o+ 328 410 360 048 230 132 .051 .008 .001 o+ 001 021 073 .205 309 .346 312 021 .001 .230 .312 346 309 205 .073 0+ .001 008 051 .132 410 328 .204 048 028 077 156 259 360 0+ 0+ .006 .590 .774 .951 0+ 0+ 002 .010 031 078 168 328 0+ 0+ 016 004 001 0+ 0+ 0+ 047 941 .735 .531 262 .118 0+ 354 393 303 187 094 037 010 .002 057 060 001 324 311 .015 O+ o+ 234 138 001 031 098 246 185 082 015 002 C N 276 312 276 002 015 082 185 324 .246 .098 031 .001 .138 .234 0+ 001 .015 060 311 0+ 303 393 354 232 057 010 037 094 187 0+ 0+ 0+ 002 016 047 118 262 531 735 941 .004 o+ D+ 001 002 0+ 0+ o+ 0+ 698 478 .210 082 028 008 932 004 0+ +0 0+ 055 017 066 372 367 247 .131 077 025 .004 o+ 0+ 0+ 002 041 124 275 318 261 .164 0+ 273 .194 097 .029 003 227 .290 0+ .004 023 .115 .194 273 290 227 .115 023 004 0+ 0+ 003 029 097 077 164 261 318 275 124 041 002 0+ 0+ .004 025 .066 on 0+ 372 257 .017 .055 131 247 367 o+ 0+ o+ 0+ 004 698 932 0+ .002 008 028 082 .210 478 0+ 0+ 0+ 0+ 0+ 0+ 0+ 0+ 004 001 923 663 430 .168 058 .017 001 +0 0+ 075 279 383 336 198 .090 031 008 041 010 .001 0+ 0+ 0+ 149 294 296 .209 109 003 047 .009 +0 124 .005 033 .147 254 279 .219 005 0-4 0+ 046 136 232 273 232 .136 .046 0+ D+ 005 033 005 0+ 0+ 009 047 .124 .219 279 254 .147 0+ o+ 296 .294 061 .003 109 209 149 o+ O+ 0+ 001 010 041 075 001 .008 090 198 383 279 031 336 +0 o+ .168 430 663 923 001 004 017 058 0+ .80 .90 .95 99 50 01 05 .10 30 40 .60 70 20 Y NOTE: 0+ represents a positive probability less than 0.0005Assume that when an adult is randomly selected, the probability that they do not require vision correction is 21%. If 9 adults are randomly selected, nd the probability that fewer than 3 of them do not require a vision correction. If 9 adults are randomly selected, the probability that fewer than 3 of them do not require a vision correction is (Round to three decimal places as needed.) In a randomized double-blind, placebo-controlled trial of children, an herb was tested as a treatment for upper respiratory infections in children. "Days of fever\" was one criterion used to measure effects. Among 350 children treated with the herb, the mean number of days with fever was 0.87, with a standard deviation of 1.53 days. Among 372 children given a placebo, the mean number of days with fever was 0.66 with a standard deviation of 1.07 days. Use a 0.01 signicance level to test the claim that the herb affects the number of days with fever. Based on these results, does the herb appear to be effective? Assume that the two samples are independent simple random samples selected from normally distributed populations, and do not assume that the population standard deviations are equal. Let population 1 be children treated with the herb. Identify the null and alternative hypotheses. [:3 A. H0: p1 # p2 C] B. H0: p1 = p2 {:3 C. H0: p1 u2 H1tu1=rt2 [:3 D. H0: p1 > p2 C] E. H0: p1 = pz {:3 F. H0: p1 = p2 H13P1=P2 H13P1 0.2 OB. Ho: P = 0.2 H1 : p = 0.2 H1: p 0.2 H1 : p = 0.2 The test statistic is z =. (Round to two decimal places as needed.) The P-value is. (Round to three decimal places as needed.) Because the P-value is the significance level, the null hypothesis. There is evidence to support the claim that the return rate is less than 20%.Listed below are annual data for various years. The data are weights (metric tons) of imported lemons and car crash fatality rates per 100,000 population. Construct a scatterplot, find the value of the linear correlation coefficient r, and find the P-value using a = 0.05. Is there sufficient evidence to conclude that there is a linear correlation between lemon imports and crash fatality rates? Do the results suggest that imported lemons cause car fatalities? Lemon Imports 232 264 357 481 532 Crash Fatality Rate 15.8 15.6 15.4 15.3 14.9 What are the null and alternative hypotheses? OA. Ho: P= 0 OB. Ho: P = 0 H1: p#0 H, : p >0 OC. Ho: P = 0 OD. Ho: p#0 H1: p