Probablity and percents

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Sue0113

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Feb 1, 2012
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In order to plan for parking and transportation needs at a college, administrators asked students how they got to school. Some came by bus, some came by car and the others used “personal “ transportation – bikes, skateboards, or walked. The following table summarizes the responses:
Gender
Male Female
Transportation
Bus 30 50
Car 37 45
Personal 19 23
What percent of the students take the bus to college?
30+50/204=.39215686X 100 = 39.21%

What percent are male or travel by car?
37+45/204=.4019 X 100= 40.19%

If a car drives up what is the probability the student is female?
37+45/45=1.8
Is this the proper way to answer these questions?
 
Sue0113 said:
In order to plan for parking and transportation needs at a college, administrators asked students how they got to school. Some came by bus, some came by car and the others used “personal “ transportation – bikes, skateboards, or walked. The following table summarizes the responses:
Gender
Male Female
Transportation
Bus 30 50
Car 37 45
Personal 19 23
What percent of the students take the bus to college?
(30+50)/204=.39215686X 100 = 39.21%

What percent are male or travel by car?
(37+45)/204=.4019 X 100= 40.19%

If a car drives up what is the probability the student is female?
(37+45)/45=1.8
Is this the proper way to answer these questions?
Click to expand...
The first is correct - but when typing inline use extra parentheses to guarantee we understand the order of operations.

For the second one, you need to add up the males, and also the cars -- but but be careful not to count the males in cars twice: (30 +37 + 19 + 45)/204

For the third one you should be very suspicious to get a number greater than 1 = probability can NEVER be >1. You have it upside-down. "Given that it is a car" means the total events you are looking at is (37+45). "Is a female" means the numerator is 45.
 
Hello, Sue0113!

In order to plan for parking and transportation needs at a college,
administrators asked students how they got to school.
Some came by bus, some came by car,
and the others used "Other" transportation – bikes, skateboards, or walked.

The following table summarizes the responses:

. . \(\displaystyle \begin{array}{c|c|c||c|} & \text{Male} & \text{Female} & \text{Total} \\ \hline \text{Bus} & 30 & 50 & 80 \\ \hline \text{Car} & 37 & 45 & 82 \\ \hline \text{Other} & 19 & 23 & 42 \\ \hline \text{Total} & 86 & 118 & 204 \\ \hline \end{array}\)

(a) What percent of the students take the bus to college?

. . \(\displaystyle \dfrac{80}{204} \:=\:0.39215686 \:\approx\:39.2\%\) . Correct!
Click to expand...


(b) What percent are male or travel by car?
Click to expand...

Answer: .\(\displaystyle \dfrac{86 + 45}{204} \:=\:\dfrac{131}{204} \:\approx\:64.2\%\)



(c) If a car drives up, what is the probability the student is female?
Click to expand...

There are 82 students who drive a car.
Of them, 45 are female.

Answer: .\(\displaystyle \dfrac{45}{82} \:\approx\:54.9\%\)
 
Thanks

Should c be answered as 54.9% or just left as the probabibility of a female driving up is .54878.
 
Sue0113 said:
Should c be answered as 54.9% or just left as the probabibility of a female driving up is .54878.
Click to expand...
% and decimal representations mean the same thing. I would call this one 0.55 or 55% - there is not enough precision in the data to justify using more than two significant digits.
 
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