Name:___________________________________
Homework #5
Due: Wed, Jun 24, 11 am
Read the directions carefully, and answer all parts of each question. Write neatly; if I can’t read it, it will
be counted wrong. You may work in groups, but you must submit your own solution. Any submissions
that are word-for-word identical will receive a score of 0. Use this sheet as a cover page or lose five
points. No late homework will be accepted under any circumstances.
1. Using the following distribution X~Uni(0,10), calculate the following probabilities:
a. Pr(X>6)
= 4*1/10 = 2/5 or 0.40
b. Pr(X<2)
=2*1/10 = 1/5 or 0.20
“The trouble with normal is that it always gets worse.”
-Bruce Cockburn
Name:___________________________________
Homework #5
Due: Wed, Jun 24, 11 am
c. Pr(X>4 or X<1)
=(1*1/10)+(6*1/10) = 7/10 or 0.70
d. Pr(3<X<5)
=2*1/10 = 1/5 or 0.2
e. Pr(X=1)
=0. The uniform is a continuous distribution therefore any X=a is 0.
2. Sketch the probabilities you calculated in #1.
“The trouble with normal is that it always gets worse.”
-Bruce Cockburn
Name:___________________________________
Homework #5
Due: Wed, Jun 24, 11 am
3. Using the following distribution X~N(10,3), n=20, calculate the following probabilities:
a. Pr (X>10)
b. Pr(X<9)
c. Pr(X>11 or X<9)
d. Pr(X>9.4)
e. Pr(9.2<X<10.2)
f. Pr(X<10.6)
g. Pr(X=9)
Pr(x>10)
Since 10 is the mean, our Z-score is 0. Thus we are interested in the area to the right of Z=0
which is simply 0.5000 by definition.
Pr(X<9)
This is the area we want, so we need the Z-value for X=9.
9 − 10
1
𝑍𝑥=9 =
= − 𝑜𝑟 − 0.33
3
3
“The trouble with normal is that it always gets worse.”
-Bruce Cockburn
Name:___________________________________
Homework #5
Due: Wed, Jun 24, 11 am
The Z-table gives us the area from the mean to the boundary, so we need the complement of
that number with respect to 0.5000. 0.5000 – 0.1293 = 0.3707.
Pr(X>11 or X<9)
This is the area we need. We got the Z-value for X=9 in the last part, so let’s see what it is for
X=11.
11 − 10 1
𝑍𝑥=11 =
= 𝑜𝑟 0.33
3
3
It’s the positive counterpart. So the two red areas are equal. All we then need to do is double
our last answer. 0.3707*2 = 0.7414
Pr(X>9.4)
So what we’re after the sum of the red and green areas. We know the green area is 0.5000
since it’s the full area to the right of the mean and the distribution is symmetric. So all we need
is the red area. Z-value first:
“The trouble with normal is that it always gets worse.”
-Bruce Cockburn
Name:___________________________________
𝑍𝑋=9.4 =
Homework #5
Due: Wed, Jun 24, 11 am
9.4 − 10 −0.6
=
= −0.2
3
3
0.0793 + 0.5000 = 0.5793.
Pr(9.2<x<10.2)
We need the sum of the red and blue areas. So, we need the corresponding Z-values. Once we
have those, we can simply read the probabilities from the table because these are tied to the
mean.
9.2 − 10
𝑍𝑥=9.2 =
= −0.27
3
10.2 − 10
𝑍𝑥=10.2 =
= 0.07
3
So we are adding 0.1064 and 0.0279 = 0.1343.
Pr(x<10.6)
“The trouble with normal is that it always gets worse.”
-Bruce Cockburn
Name:___________________________________
Homework #5
Due: Wed, Jun 24, 11 am
Once again, the sum of red and green. Again the green is 0.5000. So we need the Z-value for
10.6 and the table gives us the red.
10.6 − 10
𝑍𝑥=10.6 =
= 0.20
3
So we have 0.5000 + 0.0793 or 0.5793
Pr(x=9)
=0. Another continuous distribution.
4. Sketch the probabilities you calculated in #3.
5. Using a single sentence, explain the following:
a. Unbiasedness
The sum of the sample means equals the population mean.
b. Consistency
As we increase the sample size, the mean and standard deviation of the samples
approaches the mean and standard deviation of the population (respectively).
c. Central Limit Theorem
As we increase the sample size, our distribution becomes more normal.
“The trouble with normal is that it always gets worse.”
-Bruce Cockburn