PSY521_Week2 - Week 2 Exercise 1 6 The mean median and mode...

Info icon This preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Week 2 Exercise 1 6. The mean, median and mode all fall at the same midpoint in a normal curve. A normal curve is symmetrical. The mode is the highest point in the curve, which falls at the midpoint. The midpoint is the median value and the mean also falls at the same point. 7. The mean number is the ordinary average. It is the sum of all the scores, divided by the total number of scores. So, according to the results, the average amount of anxiety attacks experienced by all 30 participants was 6.84. SD stands for standard deviation. This is the average amount that the anxiety attacks are spread out from their average. According to these results the attacks are spread out by 3.18 attacks. Figuring out the variance and then taking the square root of that variance calculate the standard deviation. SPSS Exercises
Image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
SD= Square root of 6.60= 2.57 Exercise 2 1. School A has an average or mean of 50, which means that the total number of scores were added up and then divided by the number of scores taken. The Standard deviation is 15.25. Being that the standard deviation is the average
Image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern