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)?
1.


Chance of winning if it snows: 40%


Chance of snow: 60%


60% of 40% = (.6)(.4) = .24 = 24%





2.


Female: 62%


Females in competetive sport: 35%


35% of 62% = (.35)(.62) = .217 = 21.7%





With percent problems like this, you can do them in two parts. If you pretend your original population is 100, you can see the final result. In #2, pretend the senior class has 100 students. How many are female? It says 62%, then 62 are female. Of the 62 female students, how many plays competetive sport? The question states 35% of the females (which we now say number 62) plays a competitive sport. So, find 35% of 62 ==%26gt; (.35)(62) = 21.7%





3.


First find the mean:


(51.8 + 53.6 + 54.7 + 50.9 + 55.2 + 51.1)/6


= (317.3)/6


=52.88





S = standard deviation


S = sqrt((sum (X-M)^2)/(n-1))





X = Individual data values


M = mean of all data


n = number of data points = 6





sum (X-M)^2 = (51.8 - 52.88)^2 + (53.6 - 52.88)^2 + (54.7 - 52.88)^2 + (50.9 - 52.88)^2 + (55.2 - 52.88)^2 + (51.1 - 52.88)^2





sum (X-M)^2 = (-1.08)^2 + (.72)^2 + (1.82)^2 + (-1.98)^2 + (2.32)^2 + (-1.78)^2





sum (X-M)^2 = 17.47





S = sqrt((sum (X-M)^2)/(n-1))


S = sqrt(17.47/(6-1))


S = sqrt(17.47/5)


S = 1.87





Within S: 51.8, 53.6, 54.7, 51.1





4.


95% of all scores will fall within 2 standard deviation of the mean, that leaves 5% to fall beyond. If 1/2 score below and 1/2 score above, then 2.5% of the group should score above 96.





(240)(.025) = 6 students should score above 96.





I got the 95% value on one of the many website hits I got when I googled "standard deviation"

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