Question 1206561: A university has 14 statistics classes scheduled for its Summer 2013 term. One class has space available for 30 students, eight classes have space for 50 students, one class has space for 80 students, and four classes have space for 100 students.
(a) What is the average class size assuming each class is filled to capacity? (Enter your answer as a whole number.)
(b) Space is available for 910 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. (Enter your answers as fractions.)
x
P(X = x)
x ยท P(X = x)
30
50
80
100
(c) Find the mean of X. (Enter your answer as a whole number.)
(d) Find the standard deviation of X. (Round your answer to two decimal places.)
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
Part (a)
910 total students distributed among 14 classes
910/14 = 65 students is the average class size.
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Part (b)
Let's label the four class types as A,B,C,D
A = 30 students
B = 50 students
C = 80 students
D = 100 students
There is one copy of A and C each, 8 copies of B, and 4 copies of D.
P(A) = 1/14
P(B) = 8/14 = 4/7
P(C) = 1/14
P(D) = 4/14 = 2/7
Therefore the PDF is
| x | P(x) | | 30 | 1/14 | | 50 | 4/7 | | 80 | 1/14 | | 100 | 2/7 |
PDF = probability density function
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Part (c)
Now to fill out the x*P(x) column.
Multiply each x item with its paired P(x) probability.
Example: 30*(1/14) = 30/14 = 15/7
| x | P(x) | x*P(x) | | 30 | 1/14 | 15/7 | | 50 | 4/7 | 200/7 | | 80 | 1/14 | 40/7 | | 100 | 2/7 | 200/7 |
Adding up the items in the x*P(x) column then gets us:
mu = (15/7)+(200/7)+(40/7)+(200/7)
= (15+200+40+200)/7
= 455/7
= 65
Therefore mu = 65 is the mean or expected value.
We arrive at the same answer as part (a).
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Part (d)
We'll use mu = 65 to find the variance.
Introduce a new column called (X-mu)^2*P(X)
The naming of this column should be self-explanatory. If not then please let me know.
| X | P(X) | X*P(X) | (X-mu)^2*P(X) | | 30 | 1/14 | 15/7 | 175/2 | | 50 | 4/7 | 200/7 | 900/7 | | 80 | 1/14 | 40/7 | 225/14 | | 100 | 2/7 | 200/7 | 350 |
variance = sum of the (x-mu)^2*P(x) values
variance = (175/2)+(900/7)+(225/14)+350
variance = 4075/7
As the last set of steps:
SD = standard deviation
SD = sqrt(variance)
SD = sqrt(4075/7)
SD = 24.13 approximately when rounding to 2 decimal places.
The standard deviation is a measure of spread. The larger the standard deviation, the more spread out the distribution will be.
The same applies to the variance.
The smallest both items can be is 0, and only occurs when all values in the set are the same.
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