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MAT201 CASE 3 BIG BURGER

Big Burger Company
Sheral Wooden
Trident University
MAT201

Normal Distributions

“The final exam scores listed below are from one section of MATH 200. 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”

 (99+34+86+57+73+85+91+93+46+96+88+79+68+85+89)/15=77.933 = Mean

 ((99-77.93) ^2+(34-77.93) ^2+(86-77.93) ^2+(57-77.93)^2…..etc.)/15=362.07 = Variance(σ²)

Std Dev (σ)=√ (337.93) =19.02

77.93 ± 19.02=58.91 and 96.95

12 scores are within one standard deviation

77.93±2(19.02) =39.89 and 115.97

14 scores are within two standard deviation

“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%?”

There is a 91.15 % probability that most students will score 85% or less on the exam (or 8.85% of students will score above 85%).

0885*15 students =1.32 students will score above 85%.

Z score=(x-mean)/σ= (85-74.8)/7.57=1.34

Only 1 student.

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

To find the variance, square of the standard deviation (σ2).

“To get the best deal on a stereo system, Louis called 8 out of 20 appliance stores in his neighborhood and asked for the cost of a specific model. Below is the sample data set of prices he collected:

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

 (216+135+281+189+218+193+299+235)/8=220.75 = Mean

 ((216-220.75) ^2+(135-220.75) ^2+(281-220.75) ^2+(189-220.75) ^2+(218-220.75) ^2 etc..)/8=2731.07 = Variance(σ2)

Std Dev (σ)=√ (2731.07) =$52.26

“The Company collected a sample of the salaries of its employees. There are 70 salary data points 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.” = 6306.94

“Find the variance.” = 39777432.71

“Calculate the mean and variance of the sample data set provided below. Show and explain your steps. Round to the nearest tenth.

14, 16, 7, 9, 11, 13, 8, 10”

 (14+16+7+9+11+13+8+10)/8=11= Mean

 ((14-11)^2+(16-11)^2 etc)/8=9.7= Variance (σ2)

“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 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”

Number

Frequency

1

7

2

5

3

6

4

4

5

4

6

4

“Answer the following questions using the frequency distribution table you created”

“Which number(s) had the highest frequency?” 1

“How many times did a number of 4 or greater get thrown?” 1

“How many times was an odd number thrown?” 17

“How many times did a number greater than or equal to 2 and less than or equal to 5 get thrown?” 19

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

Range = Big Burger Co. = (175-59) = 116 sec, The Cheesy Burger = (200-79) = 121 sec

Standard deviation =Big Burger Std Dev= √ (1391.27) =37.3 sec, Cheesy Burger= √ (1260.55) =35.5 sec

Big Burger Mean = 104.88= Variance

((105-104.88) ^2+(67-104.88) ^2+(78-104.88) ^2 etc.)/8=1391.27= Variance

Cheesy Burger Mean = 128.38

 ((133-128.38) ^2+(124-128.38) ^2+(200-128.38) ^2 etc )/8=1260.55= 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?”

1 standard deviation are 68%.

2standard deviations are 95%.

3 standard deviations are 99.7%.

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