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Mod 1 Exam 1
1
...

2
...
It may be too large or too small
compared to the other values
3
...


a) How many were burgers?
b) How many were fish?
a) Burgers, 2900(0
...
28)=812

Mod 2 Exam
1
...

b) Make a relative frequency distribution for this data
...

a
...


2
...


3
...

b) Find the range of this data
...

d) Find the coefficient of variation
...


4
...

a) Is the point 57 above, below, or the same as the mean
...

b) Is the point 33 above, below, or the same as the mean
...

c) Is the point 31 above, below, or the same as the mean
...

d) Is the point 79 above, below, or the same as the mean
...

a) The data point 57 is above the mean
...
We are told that the mean is 49 and the standard deviation is 8
...

b) The data point 33 is below the mean
...
We are told that the mean is 49 and the standard deviation is 8
...

c) The data point 31 is below the mean
...
We are told that the mean is 49 and the standard deviation is 8
...
25, so the data point 31 is 2
...

d) The data point 79 is above the mean
...
We are told that the mean is 49 and the standard deviation is 8
...
75, so the data point 79 is 3
...


5
...

b) Find the mode of this set
...

Notice that there are two “middle” numbers, 15 and 18
...

Median = (15+18)/2 = 16
...

b) No number occurs more than once, so there is “no mode”
...
Find the answer to each of the following by first reducing the fractions as much as possible:
a) P(412,3)=
b) C(587,585)=

2
...

How many different passwords can you make if you cannot use any character more than once in each
password?

3
...
35, P(Bc )=
...
25
...

b) P(Ac )
...

a
...
Use P(A∪B)=P(A)+P(B)-P(A∩B)
...
So, P(B)=1-
...
55
...
35+
...
25=
...


For P(Ac )
...


P(Ac )=1-
...
65
...
For P(B)
...

P(B)=1-
...
55
...
Suppose A and B are two events with probabilities:
P(Ac )=
...
65,

a) What is (A│B) ?
b) What is (B│A) ?

P(A∩B)=
...


5
...
The company's quality control department determined that 3 % of the parts produced by
machine A, 2
...
If a part is selected at random and found to be defective, what is the probability that it was produced
by machine B?

6
...
85
...
25
...
Suppose that 7 out of 17 people are to be chosen to go on a mission trip
...


Exam 4
1
...
Suppose that you choose 7
clocks at random
...


2
...
Find P(Z ≤ 1
...

b
...
73)
...
Find P(-
...
86)
...
P(Z ≤ 1
...
89796
b
...
73= 1- P(Z ≤ -0
...
23270=0
...
P(-0
...
86)= P(Z≤0
...
09)
0
...
46414=
...
A company manufactures a large number of rods
...
0 inches and a standard deviation of
...
If you choose a rod at random, what is the
probability that the rod you chose will be:
a) Less than 3
...
7 inches?
c) Between 3
...
3 inches?

a
...
0:

So, we want P(Z ≤ -1
...
From the table, we find
...
33)=
...

b
...
7:

So, we want P(Z ≥-
...
Since this is greater than, we must use:
P(Z ≥-
...
0-P(Z ≤-
...
34458=
...

c
...
5:

and the z-score for x=4
...
67 ≤ Z ≤
...
67 ≤ Z ≤
...
4)-P( Z ≤ -
...

P(-
...
4)=
...
25143=
...

4
...
The probability
of these is given as follows:
Policies Sold
Per Day
0
1
2
3
4
5

Probability, f(x)

...
14

...
25

...
13

Find the expected number of insurance policies that the salesperson will sell per day
...


5
...
She hits the target 68% of the time
...
Suppose that you take a sample of size 20 from a population that is not normally distributed
...

2
...
In order to use the infinite
standard deviation formula, what sample size, n, should you use?
Your Answer:
n N ≤ 0
...
05
n ≤ 0
...
Suppose that in a large hospital system, that the average (mean) time that it takes for a nurse to take the
temperature and blood pressure of a patient is 150 seconds with a standard deviation of 35 seconds
...
782) on the standard normal probability distribution table
...
78) =
...

Therefore, there is a 0
...


4
...
8 % of the people have more than two jobs
...
23)
...
23)=1- P(Z<-
...
40905=
...

So there is a
...


Exam 6
1
...
You would like to know if migraine patients
prefer the old drug or the new drug
...
Find the 90% confidence limit for the
proportion of all patients that prefer the new drug
...
52

n = 190

Based on a confidence limit of 90 %, we find in table 6
...
645

So, the 90% confidence limit is:

Notice that the proportion that like the new drug may be as small as
...
5
...


2
...

You have a total of 300 patients in the village
...
Therefore, you ask 35 patients and find that 62% prefer afternoon appointments while 38% prefer morning
appointments
...


Since 62% prefer afternoon, we set P =
...
As we mentioned previously, we estimate p by P
...
62
...
A total of 35 patients were surveyed, so Based on a confidence limit of 95
%, we find in table 6
...
96
...
469 and
...


Exam 7
1
...
Also, explain what it would mean to make a
Type I error and explain what it would mean to make a Type II error
...
You believe that the newspaper’s
circulation is more than 15,000 today
...
Also, explain what it would mean to make a
Type I error and explain what it would mean to make a Type II error
...
You believe that the number of hits per month is less than
that today
...
H0: μ =15,000 circulation
H1: μ >15,000 circulation
Type I error: Reject the null hypothesis that the mean of circulations is 15,000 even though it is correct
...

b
...

Type II error: Do not reject the null hypothesis when the mean of hits per month is less than 3500
...
Suppose that we have a problem for which the null and alternative hypothesis are given by:
H0: μ=322
...

Is this a right-tailed test, left-tailed test, or two-tailed test
...
06
...
06/2=0
...
06/2=0
...
88 and z=1
...
It is recommended that pregnant women over eighteen years old get 85 milligrams of vitamin C each day
...
A doctor is concerned that her pregnant
patients are not getting enough vitamin C
...
Based on a level of significance of α =
...


H0: μ=85 milligrams per day
...

This is a left-tailed test, so we must find a z that satisfies P(Z ...
In the standard normal table, we
find z
...
05
...
05
...
05, we reject the null hypothesis
...
A mayor claims that the unemployment rate in his city is 4 %
...
So, 95 residents of the city are contacted and it is found that 8 of them are unemployed
...
02, test the hypothesis
...
04
...
04
...
z=1-0
...
98=2
...


Since this is a right-tailed test, and the z-score is greater than 2
...


Exam 8
Suppose we have independent random samples of size n1 = 420 and n2 = 510
...
38 and p2 =
...
Find the 99% confidence interval for the difference in the two population
proportions
...
Multiple choice: Which equation would you use to solve this problem?

A
...


C
...

2
...

3
...
1, we see that 99% confidence corresponds to z=2
...
Notice that the sample sizes are each
greater than 30, so we may use eqn
...
2:
B
...
-0
...
03326)
...
In certain hospital, nurses are required to constantly make rounds to check in on all of the patients
...
So, the nursing supervisor checks the
records of 70 day shift nurses and finds that they complete an average (a mean) of 30 rounds per shift with a
standard deviation of 4
...
The nursing supervisor also checks the records of 84 night shift
nurses and finds that they complete an average (a mean) of 25 rounds per shift with a standard deviation of 5
...

a) Find the 90% confidence interval for estimating the difference in the population means (µ1 - µ2)
...
Multiple choice: Which equation would you use to solve this problem?

A
...


C
...

2
...

3
...
645
...
6, s2=5
...


b) Since the entire confidence interval is positive, we can be 90 % sure that there is a difference in the
means of the two populations
...
A head librarian supervises a number of libraries in a large county
...
So, he checks the records of 40 full-time
library workers and finds that they re-shelve an average of 185 books per hour with a standard deviation of 17
...
The records of 40 part-time library show that they re-shelve an average of 190 books per hour
with a standard deviation of 9
...

Using a level of significance of α=
...
Multiple choice: Which equation would you use to solve this problem?

A
...


C
...

2
...

3
...

Since this is a two-tailed test, we must find the z that satisfies:
P(Z ...
05 and P(Z > z)=
...
05
...
645 and z=1
...
We will reject the null hypothesis if the z-score is less
than -1
...
645
...
645 and 1
...


4
...
1 7
...
3 8
...
9 7
...
4 8
...
5 8
...
1 7
...
025
...
This is a left-tailed test
...
025 = -2
...

We find the mean in the usual way:

The sample standard deviation is given by:

Then using the mean, d = -
...
2422, that we found above:

Since t < t
...

5
...
The times, in seconds, of
eight runners with and without the drink are given below:
Runner
x-time (before)
y-time (after)

1
254
265

2
276
269

3
276
277

4
265
279

5
271
266

6
273
273

Find the 95 % confidence interval for mean of the differences, µd
...
Multiple choice: Which equation would you use to solve this problem?

A
...


C
...

2
...

3
...
We will define , di = xi - yi
...
375
sd= 7
...

When we look at the student’s t chart for 95% confidence (the 95% is found along the bottom row of the
chart) and DOF=8-1=7 (the df=7 is found in the leftmost column) we find that t=2
...
Then
D
...
Suppose you have 45 data points and you calculate the sample correlation coefficient and find that r =
...
Can you
be 95% confident that a linear relation exists between the variables? If so, is the relation positive or negative? Justify you
answer
...
29396
...
32, which is above

...
So a positive linear relation exists
...
Suppose you have 60 data points and you calculate the sample correlation coefficient and find that r =
...
Can you
be 95% confident that a linear relation exists between the variables? If so, is the relation positive or negative? Justify you
answer
...
25420
...
20, which is below

...
So no linear relation exists
...
Compute the sample correlation coefficient for the following data:

Can you be 95% confident that a linear relation exists between the variables? If so, is the relation positive or
negative? Justify you answer
...
966

IrI=0
...
87834

x*=6
...
702

Sy=3
...
There is a negative
linear relation between the variables
...
Find the best fit line for the following data:

r =
...
2 Sy = 5
...


y= 1
...
123
...
005 in the right tail of the chi-square
distribution
...
005, then look down the left column for 17
...
718
...
Find the value of X2 for 10 degrees of freedom and an area of
...

Since the chi-square distribution table gives the area in the right tail, we must use 1 -
...
995
...
995, then look down the left column for 10
...
156
...
Find the value of X2 values that separate the middle 80 % from the rest of the distribution for 9 degrees of freedom
...
80=
...
20/2 =
...

Notice that the area to the right of the first X2 is
...
10 =
...
So we use this value and a DOF of 9 to get
X2 = 4
...

The area to the right of the second X2 is
...
So we use this value and a DOF of 9 to get X2 = 14
...

4
...
05
In order to solve this, we turn to the F distribution table that an area of
...
DOF=(4,17) indicates that degrees of
freedom for the numerator is 4 and degrees of freedom for the denominator is 17
...
96
...
The mayor of a large city claims that 30 % of the families in the city earn more than $ 100,000 per year; 52
% earn between $ 30,000 and $ 100,000 (inclusive); 18 % earn less than $ 30,000 per year
...

Test the mayor’s claim based on 5 % significance level
...

H1: The mayor’s distribution is not correct
...

Calculate the degrees of freedom for three possible outcomes: DOF=3-1=2
...
05)
...
05 on the Chi-square distribution table to get 5
...

For the expected frequencies, we will use Ei = npi
...
991
...


6
...
So, they give a
test for alertness to two groups of drivers
...
The results of
the tests are given below
...
) Test at the 1 %
level of significance
...

H1: Driving hours and alertness are not independent events
...
The degrees of freedom are
given by:
DOF = (# of Rows-1)(# of Columns-1)=(2-1)(2-1)=1
...
01 (for the 1% level of
significance) we find in the chi-square table a critical value of 6
...


This value is greater than the critical value of 6
...
So, we reject the null hypothesis

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Title: MATH 110 All Exams STATISTICS (LATEST UPDATE ) 20222023- Portage Learning
Description: MATH 110 All Exams STATISTICS (LATEST UPDATE ) 20222023- Portage Learning

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