Student loan debt is the only form of consumer debt that has grown since the peak of consumer debt in 2008. The average student loan of somebody younger than 30 is $20,550. Assume the standard deviation for debt is $4,500 per student. a. What is the probability that the sample mean will be less than $19,000 for a sample size of 30 students? b. Identify the symmetrical interval that includes 93% of the sample means if the true population mean is $20,550 per student. c. Answer the question in part a for a sample size of 60. Explain the differences in these two probabilities.
Student loan debt is the only form of consumer debt that has grown since the peak of consumer debt in 2008. The average student loan of somebody younger than 30 is $20,550. Assume the standard deviation for debt is $4,500 per student. a. What is the probability that the sample mean will be less than $19,000 for a sample size of 30 students? b. Identify the symmetrical interval that includes 93% of the sample means if the true population mean is $20,550 per student. c. Answer the question in part a for a sample size of 60. Explain the differences in these two probabilities.
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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Student loan debt is the only form of consumer debt that has grown since the peak of consumer debt in 2008. The average student loan of somebody younger than 30 is

Transcribed Image Text:**Student Loan Debt Analysis**
Student loan debt has uniquely continued to grow since the 2008 peak of consumer debt. Currently, the average student loan for individuals under 30 amounts to $20,550, with a standard deviation of $4,500 per student.
**Tasks:**
**a. Probability Analysis:**
Calculate the probability that the sample mean will be less than $19,000 for a sample size of 30 students.
**b. Confidence Interval:**
Determine the symmetrical interval that covers 93% of the sample means if the true population mean is $20,550 per student.
**c. Comparative Probability:**
Re-evaluate the probability scenario in part a for a sample size of 60 students and explain the difference in probabilities between the two sample sizes.
Expert Solution

Step 1
a.
The given mean is $20,500, and standard deviation is $4,500.
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