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Case Frequency Distributions

Math Problem Instructions:

Case Assignment

By submitting this assignment, you affirm that it contains all original work, and that you are familiar with Trident University’s Academic Integrity policy in the Trident Policy Handbook. You affirm that you have not engaged in direct duplication, copy/pasting, sharing assignments, collaboration with others, contract cheating and/or obtaining answers online, paraphrasing, or submitting/facilitating the submission of prior work. Work found to be unoriginal and in violation of this policy is subject to consequences such as a failing grade on the assignment, a failing grade in the course, and/or elevated academic sanctions. You affirm that the assignment was completed individually, and all work presented is your own.





Problems need to include all required steps and answer(s) for full credit. All answers need to be reduced to lowest terms where possible.



Answer the following problems showing your work and explaining (or analyzing) your results.



The math grades on the final exam varied greatly. Using the scores below, how many scores were within one standard deviation of the mean? How many scores were within two standard deviations of the mean?

99 34 86 57 73 85 91 93 46 96 88 79 68 85 89



The scores for math test #3 were normally distributed. If 15 students had a mean score of 74.8% and a standard deviation of 7.57, how many students scored above an 85%?

If you know the standard deviation, how do you find the variance?

To get the best deal on a stereo system, Louis called eight appliance stores and asked for the cost of a specific model. The prices he was quoted are listed below:

$216 $135 $281 $189 $218 $193 $299 $235



Find the standard deviation.



A company has 70 employees whose salaries are summarized in the frequency distribution below.

Salary Number of Employees

5,001–10,000 8

10,001–15,000 12

15,001–20,000 20

20,001–25,000 17

25,001–30,000 13

Find the standard deviation.

Find the variance.

6. Calculate the mean and variance of the data. Show and explain your steps. Round to the nearest tenth.



14, 16, 7, 9, 11, 13, 8, 10



Create a frequency distribution table for the number of times a number was rolled on a die. (It may be helpful to print or write out all of the numbers so none are excluded.)

3, 5, 1, 6, 1, 2, 2, 6, 3, 4, 5, 1, 1, 3, 4, 2, 1, 6, 5, 3, 4, 2, 1, 3, 2, 4, 6, 5, 3, 1



Answer the following questions using the frequency distribution table you created in No. 7.

Which number(s) had the highest frequency?

How many times did a number of 4 or greater get thrown?

How many times was an odd number thrown?

How many times did a number greater than or equal to 2 and less than or equal to 5 get thrown?

The wait times (in seconds) for fast food service at two burger companies were recorded for quality assurance. Using the data below, find the following for each sample.

Range

Standard deviation

Variance

Lastly, compare the two sets of results.



Company Wait times in seconds

Big Burger Company 105 67 78 120 175 115 120 59

The Cheesy Burger 133 124 200 79 101 147 118 125

What does it mean if a graph is normally distributed? What percent of values fall within 1, 2, and 3, standard deviations from the mean?

Math Problem Sample Content Preview:

Module 3 – Case Frequency Distributions
Name:
Institution:
Response to question 1
Math grades: 99 34 86 57 73 85 91 93 46 96 88 79 68 85 89
Mean
Mean = Sum of math grades (X) / N (Number of students)
Mean= (99+34+86+57+73+85+91+93+46+96+88+79+68+85+89)/15
Mean=77.93333
Standard Deviation
To get standard deviation we find the square root of variance
S=‑X-M2n-1
Step 1: calculation of variance
Subtract the mean grade from each of the values
99-77.93333=21.06667
34-77.93333=-43.93333
86-77.93333=8.06667
57-77.93333=-20.93333
73-77.93333=-4.93333
85-77.93333=7.06667
91-77.93333=13.06667
93-77.93333=15.06667
46-77.93333=-31.93333
96-77.93333=18.06667
88-77.93333=10.06667
79-77.93333=1.06667
68-77.93333=-9.93333
85-77.93333=7.06667
89-77.93333=11.06667
Next you square all answers from subtraction above
(21.06667)2=443.80458
(-43.93333)2=1930.13748
(8.06667)2=65.07116
(-20.93333)2=438.20430
(-4.93333)2=24.33774
(7.06667)2=49.93782
(13.06667)2=170.73786
(15.06667)2=227.00454
(-31.93333)2=1019.73756
(18.06667)2=326.40456
(10.06667)2=101.33784
(1.06667)2=1.13778
(-9.93333)2=98.67104
(7.06667)2=49.93782
(11.06667)2=122.47118
Then add all the squared numbers
(443.80458+1930.13748+65.07116+438.20430+24.33774+49.93782+170.73786+227.00454+
1019.73756+326.40456+101.33784+1.13778+98.67104+49.93782+122.47118=5068.93326
Then divide the sum of squares by (n-1)
Where n=15, n-1 is 15-1=14
5068.93326/14=362.06666
Variance=362.06666
Step 2 calculate standard deviation
Standard deviation is the square root of variance
Standard deviation (S)= 362.06666
S=19.02805
* Scores within one standard deviation of the mean
Mean=77.93333
To get scores within one standard deviation 77.93333 +1 /-1
77.93333-1=76.93333
77.93333+1=78.93333
There is no score within one standard deviation of the mean. The nearest score is 79 but falls outside the range of 76.93333 and 78.93333
* Scores within two standard deviations of the mean
To get scores within two standard deviations 77.93333 +2 /-2
77.93333-2=75.9333
77.93333+2=79.93333
There is only one score within two standard deviations of the mean that is 79
Response to question 2
Mean=74.8%
Standard deviation=7.57
* Approximation of students who scored above 85%
Mean score=74.8%
Standard deviation of the marks =7.57
First step: Z-score for 85%
z = (X - μ) / σ
(85%-74.8%)/7.57=1.35
Second step: P (Z<1.35) =0.9115
Third step: subtract this from 1
1-0.9115=0.0885
Probability that some students scored above 85% is 0.0885
Students who scored above 85%
0.0885*15=1.3275
Round off since there is no fraction of a student
Two students scored above 85%.
* finding variance when given the standard deviation
If you are given the standard deviation, you simply square it to get the variance
Standard deviation2=variance
Variance=7.572
Variance=57.3049
Response to question 3
Standard Deviation
To get standard deviation we find the square root of variance
S=‑X-M2n-1
Step 1: calculation of variance
Subtract the mean grade from each of the values
Mean = ($216+ $135+$281+ $189+ $218+ $193+ $299+ $235)/8
Mean= $220.75
$216-$220.75=-$4.75
$135-$220.75=-$85.75
$281-$220.75=$60.25
$189-$220.75=-$31.75
$218-$220.75=-$2.75
$193-$220.75=-$27.75
$299-$220.75=$78.25
$235-$220.75=$14.25
Next you square all answers from subtraction above
(-$4.75)2=$22.5625
(-$85.75)2=$7353.0625
($60.25)2=$3630.0625
(-$31.75)2=$1008.0625
(-$2.75)2=$7.5625
(-$27.75)2=$770.0625
($78.25)2=$6123.0625
($14.25)2=$203.0625
Then add all the squared numbers
($22.5625+$7353.0625+$3630.0625+$1008.0625+$7.5625+$770.0625+$6123.0625+$203.062)=$19117.5
Then divide the sum of squares by (n-1)
n=8, n-1 is 8-1=7
19117.5/7=2731.07143
Variance=2731.07143
Step 2 calculate standard deviation
Standard deviation is the square root of variance
Standard deviation (S) = 2731.07143
S=$52.25965
Response to question 4
Step 1 identify the lower and upper limit
Class

Frequency

Lower limit

Upper limit

5,001-10,000

8

5,001

10,000

10,001-15,000

12

10,001

15,000

15,001-20,000

20

15,001

20,000

20,001-25,000

17

20,001

25,000

25,001-30,000

13

25,001

30,000

Step 2 use the average of lower and upper-class limit to get the mid-point
Class

Frequency

Lower limit

Upper limit

Mid-point

5,001-10,000

8

5,001

10,000

7,500.5

10,001-15,000

12

10,001

15,000

12,500.5

15,001-20,000

20

15,001

20,000

17,500.5

20,001-25,000

17

20,001

25,000

22,500.5

25,001-30,000

13

25,001

30,000

27,500.5

Step 3 get the mean using the formulae:
μ=‑f⋅M/‑f
μ= is the mean
‑f⋅M=is the sum of frequency multiplied by the midpoint
‑f= summation of frequency
Class

Frequency

Mid-point(M)

f.M

M2

f.M2

5,001-10,000

8

7,500.5

60,004

56,257,500.25

450,060,002

10,001-15,000


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