SOLUTION: Dear tutor,
Please help me with this homework...I appreciate it!!
Florida State University has 14 statistics classes scheduled for its Summer 2013 term. One class has space a
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Question 1001120: Dear tutor,
Please help me with this homework...I appreciate it!!
Florida State University has 14 statistics classes scheduled for its Summer 2013 term. One class has space available for 30 students, eight classes have space for 60 students, one has space for 70 students, and four classes have space for 100 students.
a. What is the average class size assuming each class is filled to capacity ?
980/14=70
b. Space is available for 980 students. Suppose that each class is filled to capacity and select a statistics student at random. Let the random variable X equal the size of the student’s class. Define the PDF for X.
x p(x) X*P(X)
30 1/14 2.14
60 8/14 34.29
70 1/14 5
100 4/14 28.57
I really have a hard time with this PDF.
c. Find the mean of X.
Add 2.14 34.39 5 and 28.57 =70
d. Find the standard deviation of X.
(x-mean)^2 x p(x) for each value and add them and do square root of the sum.
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
a)
Info: "One class has space available for 30 students, eight classes have space for 60 students, one has space for 70 students, and four classes have space for 100 students. "
Compute the weighted mean based on the info
1*30 + 8*60 + 1*70 + 4*100 = 980
980/(1+8+1+4) = 980/14 = 70
Average number of students per class is 70 students. You have the correct answer.
=====================================================
b)
X = number of students in a single class
| X | P(X) |
|---|
| 30 | 1/14 |
| 60 | 8/14 = 4/7 |
| 70 | 1/14 |
| 100 | 4/14 = 2/7 |
The PDF is simply the table shown above. It's the set of all possible probabilities for every scenario.
Take note how there are only 2 columns in this PDF. The input column X and the output column P(X)
=====================================================
c)
Now we introduce the X*P(X) column
| X | P(X) | X*P(X) |
|---|
| 30 | 1/14 | 2.142857 |
| 60 | 4/7 | 34.28571 |
| 70 | 1/14 | 5 |
| 100 | 2/7 | 28.571429 |
sum the values in the last column
2.142857+34.28571+5+28.571429 = 69.999996
Due to rounding error, we don't get exactly 70 if we use the decimal values. If you stick with fractions the whole time, then you'll get 70 exactly
Either way, the answer is 70 like in part (a)
=====================================================
d)
from parts (a) and (c), we know the mean is mu = 70
Now add on the columns X - mu, (X-mu)^2, and (X - mu)^2*P(X)
| X | P(X) | X*P(X) | X - mu | (X - mu)^2 | (X - mu)^2*P(X) |
|---|
| 30 | 1/14 | 2.142857143 | -40 | 1600 | 114.2857143 |
| 60 | 4/7 | 34.28571429 | -10 | 100 | 57.14285714 |
| 70 | 1/14 | 5 | 0 | 0 | 0 |
| 100 | 2/7 | 28.57142857 | 30 | 900 | 257.1428571 |
Add up the values in the last column:
114.2857143 + 57.14285714 + 0 + 257.1428571 = 428.57142854
The variance of X is approximately 428.57142854
Then take the square root of the variance to get sqrt(428.57142854) = 20.7019667795116
So the standard deviation of X is approximately 20.7019667795116
Here is a page that offers another example on finding the standard deviation:
https://www.ltcconline.net/greenl/courses/201/probdist/random.htm
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