Download Solution to Test # 1 - University of Arizona Math

Math 115 B - Section 4 - Test 1
Name:___________________
09 October 2000
In order to get full credit, you must show all of the steps and all of the work that you
use to solve each problem.
A list of formulas that you may find useful is provided on the last page of this test.
Question
Points
1
14
2
14
3
14
4
20
5
16
6
22
Total
100
Score
1.
The table shown below was obtained by running the "Histogram" function of
Excel on a random sample of size n=100. This random sample consists of
independent observations of a continuous random variable X.
Bin
3.5
4
4.5
5
5.5
6
6.5
7
More
Total
Relative Adjusted
Frequency frequencies heights
1
0.01
0.02
2
0.02
0.04
19
0.19
0.38
42
0.42
0.84
21
0.21
0.42
12
0.12
0.24
3
0.03
0.06
0
0
0
0
0
0
100
1
2
a. (10 points) Use this information to plot an approximation of the p.d.f. for
the random variable X.
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
3.5
4
4.5
5
5.5
6
6.5
7
x
b. (4 points) Use the plot found in part (a) to estimate the mean and standard
deviation of X. Explain.
Since the graph above is roughly symmetric, the expected value of X is
close to the value at which the p.d.f. is maximum, that is X  5, and the
standard deviation is about X  0.5.
2. The plot below shows the graph of the function f(x)=9x-x3 for x in the interval
[0,3]. We want to use midpoint sums to compute the area under the curve of f for
x between 0 and 3.
12
10
f(x)
8
6
4
2
0
0
0.5
1
1.5
2
2.5
3
3.5
x
a. (2 points) Find the points x0, x1, x2, and x3, which subdivide the interval
[0,3] into three subintervals of equal length.
x0 = 0
x1 = 1
x2 = 2
x3 = 3
b. (2 points) Find the midpoints m1, m2 and m3, of each of the subintervals.
What is the width, x, of each subinterval ?
m0 = 0.5
m1 = 1.5
m2 = 2.5
x = 1
c. (10 points) Find the midpoint sum S3(f, [0, 3]).
3
S 3 ( f , [0,3])   f (mi )  x   f (0.5)  f (1.5)  f (2.5)  1
i 1
 4.375  10.125  6.875  21.375
3. Consider a discrete random variable X which can take the values 0, 1, 2, 3, and 4,
with the probabilities given below.
x
P(X=x)
0
0.3
1
0.2
2
0.1
3
0.15
4
0.25
a. (3 points) Find the mean of X
 X   x  P( X  x)  0  0.3  1  0.2  2  0.1  3  0.15  4  0.25
all x
 1.85
b. (5 points) Find the standard deviation of X
V ( X )   ( x   X ) 2  P( X  x)
all x
 (0  1.85) 2  0.3  (1  1.85) 2  0.2  (2  1.85) 2  0.1  (3  1.85) 2  0.15  (4  1.85) 2  0.25
 2.5275
 X  V ( X )  2.5275  1.59
c. (6 points) Recall that the c.d.f. of X is defined by FX(x)=P(X x). Find the
values of FX(x) for x = 2, 4, and 5.
FX(2) = P(X=0) + P(X=1) + P(X=2) = 0.3 + 0.2 + 0.1 = 0.6
FX(4) = FX(2) + P(X=3) + P(X=4) = 0.6 + 0.15 + 0.25 = 1.0
FX(5) = FX(4) = 1.0
4. Consider the continuous random variable X which takes values in [0,) and
whose p.d.f. fX(x)= x exp(-x) is plotted below.
0.4
0.35
p.d.f. of x
0.3
0.25
0.2
0.15
0.1
0.05
0
0
2
4
6
8
10
12
x
a. (6 points) On the graph above, shade the region of the plot that
corresponds to P(2 X  6). Give a rough estimate of this probability.
The area of the shaded region can be estimated by first computing the area
under the graph of fX for x between 2 and 4. This area is close to that of a
parallelogram of base 2 and sides of length 0.26 and 0.07 respectively.
This gives an approximate value of 0.33 for P(2 X  4). The probability
P(4 X  6) can be estimated by computing the area of a triangle of base 2
and height 0.07. This area is equal to 0.07. An estimate of P(2 X  6) is
thus 0.33 + 0.07 = 0.40.
b. (4 points) Set up, but do not evaluate, an integral that corresponds to
P(2 X  6).
6 
P(2  X  6) 
  x  e  dx
 
x
2
c. (6 points) The c.d.f. of X is given by FX (x) = -(1+x) exp(-x). Use this
information to find P(2 X  6). Is this result in agreement with the
estimate found in (a) ?
P(2  X  6)  FX (6)  FX (2)  7  e 6  3  e 2  0.39
This result is in good agreement with the area estimated in (a).
d. (4 points) Set up, but do not evaluate, an integral that gives the expected
value of X.

 X    x 2  e -x  dx
0
5. (4 points each) For each of the plots below, indicate whether the function that is
represented could be the p.d.f. or the c.d.f. of a continuous random variable. If
you think that it is neither a p.d.f. nor a c.d.f., circle “none”. If you think it is the
plot of a p.d.f., indicate whether the distribution is uniform, exponential, normal
or standard normal. In each case, give brief explanation of your reasoning.
a. Plot a (circle all that apply)
p.d.f.
uniform
exponential
normal
standard normal
c.d.f.
None
Plot a
1.2
1
0.8
0.6
0.4
0.2
0
Explanation: The function is always
Increasing; it is 0 for large negative values of
x and 1 for large positive values of x.
b. Plot b (circle all that apply)
p.d.f.
uniform
exponential
normal
standard normal
c.d.f.
None
0
2
4
6
8
10
x
Plot b
0.6
0.5
0.4
0.3
0.2
0.1
Explanation: The area under the curve is
0.50.3 = 1.5 > 1, so it cannot be the graph
of a p.d.f. Since the function is not always
increasing, it cannot be the graph of a c.d.f.
c. Plot c (circle all that apply)
p.d.f.
uniform
exponential
normal
standard normal
c.d.f.
None
0
0
2
4
6
8
10
6
8
10
x
Plot c
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0
2
4
x
Explanation: This looks like the graph of the p.d.f. of an exponential random
variable with mean   1.5
d. Plot d (circle all that apply)
p.d.f.
uniform
exponential
normal
standard normal
c.d.f.
None
Plot d
0.6
0.5
0.4
0.3
0.2
0.1
0
Explanation: This graph looks like the p.d.f.
of a normal random variable with mean 5 and
standard deviation X  0.8.
0
2
4
6
8
x
6. Consider an oil tract of value Vt = $ 220,000,000. Assume that we ask geologists
to estimate the value of the tract before its proven value is known. We can
consider that their estimates are observations of a normal random variable X of
mean X = Vt and of standard deviation X = $ 25,000,000.
a. (8 points) The following numbers are estimates, in millions of dollars,
given by four geologists. They form a random sample of size n = 4:
x1 = 223.47, x2 = 274.09, x3 = 217.89, x4 = 213.59. Compute the
corresponding sample mean, sample variance and sample standard
deviation.
1 n
1
x   xi  223.47  274.09  217.89  213.59 
n i 1
4
 232.26 million dollars
s2 
1 n
 x i  x 2

n  1 i 1
2
2
1  223.47  232.26   274.09  232.26  
 
3   217.89  232.26 2  213.59  232.26 2 
 794.03 (million dollars) 2
s  s 2  28.18 million dollars
b. (8 points) Assume that we get estimates from 100 geologists. If we call x
the corresponding sample mean, find the expected value and standard
deviation of x .
10
x 
E ( x )  E ( X )  $220,000,000
X
n

25000000
 $25,000,000
10
c. (6 points) Give a formula for the standardization S of x . What can you
say about the distribution of S ?
S
x  x
x

x  Vt
X
 10
x  220,000,000
25,000,000
10
The distribution of S is standard normal. It is therefore given by the
2
1
following formula f x ( s) 
 e s / 2
2
FORMULAS
1
 e 0.5z
2 
f Z ( z) 
2

E ( X )   x  f X ( x)
E ( X )   x  f X ( x) dx

all x
V ( X )    x   X   f X ( x)
V (X ) 
2

 x   
2
X
 f X ( x) dx

all x
 X2  V ( X )
ba
n
xi  a  i  x
1 n
  xi
n i 1
s2 
x 
x
S
X  X
X

mi 

n
1
  xi  x
n  1 i 1
V x 
V (X )
n

2
xi 1  xi
2
S n  f , [a, b]   f (mi )  x
n
i 1