6 3 homework help mat 133 - Study guides, Class notes & Summaries

MAT 133-J4254 | MAT 133 6-3 Homework Help | Assignment Solutions 2023 (MATH) Students who score in the top 7 percent are recognized publicly for their achievement by the Department of the Treasury. Assuming a normal distribution, how many standard deviations above the mean does a student have to score to be publicly recognized? Normal Distribution: Top 7% is the same as the bottom 93% Using Excel you would use =NORM.S.INV(p) where p is the area to the left of the z-score. =NORM.S.INV(.93...

6-3 Homework Help MAT 133 | Assignment Solutions 2023. Joan’s finishing time for the Bolder Boulder 10K race was 1.75 standard deviations faster than the women’s average for her age group. There were 405 women who ran in her age group. Assuming a normal distribution, how many women ran faster than Joan? We need to figure out what percent (technically, proportion) of women had a shorter run time. Then we multiply that by 405 to figure out the number of women that were faster. If Joan...

4-3 Homework Help MAT 133-J4254 | Questions with Solutions. Suppose the probability of an IRS audit is 1.4 percent for U.S. taxpayers who file form 1040 and who earned $100,000 or more. (a) What are the odds that such a taxpayer will be audited? P(A) = P(the taxpayer will be audited): 1.4% → .014 P(A’) = P(the taxpayer won’t be audited) = 1 - .014 = .986 b) What are the odds against such a taxpayer being audited? Here is a video you can watch for Problem 2: Homework Problem 3 Reca...

MAT 133 4-3 Homework Help | Questions with Solutions. So you need to multiply your individual probabilities together to see if it equals the probability of them both happening. Which pairs of events are independent? P(J) = .50, P(K) = .40, P(J ∩ K) = .3. P(J) ×P(K) = .5 ×.4 =.2 and P(J ∩ K) = .3, therefore J and K are not independent. P(J) = .60, P(K) = .20, P(J ∩ K) = .12. P(J) × P(K) = .6 ×.2 =.12 and P(J ∩ K) = .12, therefore J and K are independent. P(J) = .15, P(K) = .5,...