# Arithmetic Mean and Students Essay

ANSWER: Predicted GPA = 3. 45 From StatCrunch: [pic] 2. Using the MM207 Student Data Set: a) Select a continuous variable that you suspect would not follow a normal distribution.

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ANSWER: Continuous Variable: Age b) Create a graph for the variable you have selected to show its distribution. ANSWER: [pic] c) Explain why these data might not be normally distributed. ANSWER: The reason why I chose age as an irregular distributed variable is because there are all different age groups that still study today.As online is getting more and more practical for students, individuals that are older are taking advantage of graduating and studying collegiately. d) Select a second continuous variable that you believe would approximate a normal distribution ANSWER: Height e) Create a graph to show its distribution.

ANSWER: [pic] f) Explain why these data might be normally distributed. .ANSWER: Initially, I thought that height would not have a great deal of frequency. I was proven otherwise when I created the frequency chart for all the heights. As you can see, larger heights are more compacted together in the middle near the mean. . Jonathan is a 42 year old male student and Mary is a 37 year old female student thinking about taking this class. Based on their relative position, which student would be farther away from the average age of their gender group based on this sample of MM207 students? ANSWER: Johnathan From StatCrunch: [pic] 4.

If you were to randomly select a student from the set of students who have completed the survey, what is the probability that you would select a male? Explain your answer. ANSWER: Probability: 0. 165 Converted to a Percentage of 16. 5% Females: 85 Males: 17Didn’t Answer: 1 Total # of Students: 103 Formula Used: (total # of males) / (total # of students (M & F) + total # unanswered) = Probability of Males Plugging in Numbers: 17 / (102 + 1) ( 0. 165 *(Rounded to the nearest thousandth) Converted to a percentage of 16. 5% From StatCrunch: [pic] 5. Using the sample of MM207 students: a) What is the probability of randomly selecting a person who is conservative and then selecting from that group someone who is a nursing major? ANSWER: Total # of Students that answered: Conservative ( 26%Conservative Nursing ( 11%Total # of Students (including those that didn’t answer): Conservative ( 25%Conservative Nursing ( 11% * Based on the 99 students that answered the question: Conservative: 26/99 = 0. 2626 ( 26% (rounded to the nearest percentage) Conservative Nursing: 11/99 = 0.

1111 ( 11% (rounded to the nearest percentage) * Based on the total number of students (including the 4 that didn’t answer): Conservative: 26/103 = 0. 2524 ( 25% (rounded to the nearest percentage) Conservative Nursing: 11/103 = 0. 1068 ( 11% (rounded to the nearest percentage) From StatCrunch: [pic] From StatCrunch: pic] b) What is the probability of randomly selecting a liberal or a male? ANSWER: 98 Students: 0. 3878 or 38. 78% 103 Students: 0. 3689 or 36. 89% 98 total students who answered the question: Probability of selecting a liberal: 23/98 (98 = students who answered the question) = 0.

2345 or 23. 45% (rounded to the nearest hundredth) Probability of selecting a male that is liberal/moderate/conservative = 17/98 or 17. 35% Subtracted by the number of individuals that are male and liberal (-2) Formula: 23 + 17 – 2 = 38/98 = 0. 38775510204 or 38. 78% (rounded to the nearest hundredth) 03 total students who were part of the survey: Probability of selecting a liberal: 23/98 (98 = students who answered the question) = 0. 2345 or 23. 45% (rounded to the nearest hundredth) Probability of selecting a male that is liberal/moderate/conservative = 17/98 or 17. 35% Subtracted by the number of individuals that are male and liberal (-2) Formula: 23 + 17 – 2 = 38/103 = 0.

36893203883 or 36. 89% (rounded to the nearest hundredth) From StatCrunch: [pic]6. Facebook reports that the average number of Facebook friends worldwide is 175. 5 with a standard deviation of 90. 7. If you were to take a sample of 25 students, what is the probability that the mean number Facebook friends in the sample will be 190 friends or more? ANSWER: 0.

2119 Mean: 175. 5 Standard Deviation: 90. 57 Sample: 25 Probability Formula Used: P(X&gt;190) = P((X-mean)/sqrt(n)) Plugging in the numbers: (190-175. 5) / (90. 57/5) 14. 5 / 18. 114 = 0.

800* (rounded to the nearest thousandth) Z- Score Correspondence: -0. 8 under the Z of 0. 00 correlates to . 2119 7. Select a random sample of 35 student responses to question 6, “How many credit hours are you taking this term? Using the information from this sample, and assuming that our data set is a random sample of all Kaplan statistics students, estimate the average number of credit hours that all Kaplan statistics students are taking this term using a 95% level of confidence. Be sure to show the data from your sample and the data to support your estimate.

ANSWER: Lower Limit: 9. 84 Upper Limit: 11. 83 Mean = 10. 83333333 Sample = (11,12,10,11,11,15,11,11,11,11,10,12,16,10,12,10,6,6,12,10,4,15,11,12,10,3,10,12,12,18) Standard Deviation = 2.

756078 (Refer to StatCrunch Below) Std. Error = 0. 8213584 95% Confidence Interval Data: ? : population mean Standard Deviation: 2. 7556078 N = 30 Sample Mean: 10. 833333 Lower Limit: 9. 84 Upper Limit: 11.

83 From StatCrunch: [pic] 8. Assume that the MM207 Student Data Set is a random sample of all Kaplan students; estimate the proportion of all Kaplan students who are male using a 90% level of confidence. ANSWER: Lower = 0. 105 Upper = 0. 225 Frequency Table Results for Gender: (# of male students) / (# of total students) 17 / 103 = 0. 165 1 – 0.

65 = 0. 835 sqrt (0. 165*(0.

835/103)) = sqrt(0. 165*0. 0081067961) = sqrt(0. 0013376214) =0.

03657350680479 Z = 90% or 1. 645 Margin of Error = 1. 645 * 0. 03657350680479 =0. 060163441868 Lower = 0. 165 – 0. 060163441868 = 0.

10483655814 Upper = 0. 165 + 0. 060163441868 = 0. 22516344186 Lower = 0. 105 Upper = 0. 225 From StatCrunch: [pic] Total # of students who didn’t answer = 1 Total # of students = 103 9.

Assume you want to estimate with the proportion of students who commute less than 5 miles to work within 2%, what sample size would you need?ANSWER: 2,500 is the minimum sample size needed N = 1 E^2 = 1 / (0. 02)^2 = 1/0. 004 = 2,500 10. A professor at Kaplan University claims that the average age of all Kaplan students is 36 years old. Use a 95% confidence interval to test the professor’s claim. Is the professor’s claim reasonable or not? Explain. ANSWER: I would have to reject the professor’s prediction of the average age of Kaplan students to be 36 years old. At 95% confidence interval, the results show the range to be 37.

25-41. 31. Since 36

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