Math Question (4 Questions)?

1.) A football team has a 60% chance of winning when it doesn't snow but only a 40% chance of winning if it does snow. Suppose there is a 60% chance of snow. WHAT IS THE PROBABILITY THAT THE TEAM WILL WIN?





2.) The senior class is 62% female. 35% of the females play a competitive sport. FIND THE PROBABILITY THAT A STUDENT PLAYS A COMPETITIVE SPORT, GIVEN THE STUDENT IS FEMALE?





3.) The daily power usage for 6 days in June were: 51.8, 53.6, 54.7, 50.9, 55.2, and 51.1. FIND THE STANDARD DEVIATION FOR THESE DATA. WHICH DATA FALL WITHIN STANDARD DEVIATION FROM THE MEAN?





4.) Scores on an exam are normally distributed with a mean of 76 and a standard deviation of 10. In a group of 240 students, HOW MANY STUDENTS SCORE ABOVE 96?

Math Question (4 Questions)?
Hi,





1.) A football team has a 60% chance of winning when it doesn't snow but only a 40% chance of winning if it does snow. Suppose there is a 60% chance of snow. WHAT IS THE PROBABILITY THAT THE TEAM WILL WIN?





48%





60% snow x 40% win = 24% chance of winning


40% no snow x 60% win = 24% chance of winning


24% + 24% = 48% chance of winning








2.) The senior class is 62% female. 35% of the females play a competitive sport. FIND THE PROBABILITY THAT A STUDENT PLAYS A COMPETITIVE SPORT, GIVEN THE STUDENT IS FEMALE?





If you are looking for the probability that a senior is an athlete and a female, that is 62% x 35% = 21.7%





If you are looking for the probability that a senior is an athlete given that you are limited to considering only females, then it is 35% that she plays a competitive sport.











3.) The daily power usage for 6 days in June were: 51.8, 53.6, 54.7, 50.9, 55.2, and 51.1. FIND THE STANDARD DEVIATION FOR THESE DATA. WHICH DATA FALL WITHIN STANDARD DEVIATION FROM THE MEAN?





The mean of these values is 52.88333.





The standard deviation is √((51.8 - 52.88333)² + (53.6 - 52.88333)² + (54.7 - 52.88333)² + (50.9 - 52.88333)² + (55.2 - 52.88333)² + (51.1 - 52.88333)²) = √17.4683 = 4.18.





All 6 fall within one standard deviation of the mean.








4.) Scores on an exam are normally distributed with a mean of 76 and a standard deviation of 10. In a group of 240 students, HOW MANY STUDENTS SCORE ABOVE 96?





normalcdf(96,999,76,10) = .02275 = 2.275%





I hope that helps!! :-)
Reply:1) 60/100 (does snow) * 60/100 (winning) = 36/100


2) 100/100 (female) * 35/100 (plays) = 35/100


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