Stats_Material43 - Q1 Ans A i Mean = Sum of the...

Info icon This preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
Q1 Ans. A. i.  Mean = Sum of the observations / No.of the observations Here n = 20 ; Sum of the observations = 127 Mean = 127/20 = 6.35 ii.  Median is the middle observation. If N is odd, use the formula (N+1)/2 th observation. If it is even use the formula[N/2th observation + (N/2 +1)th observation]/2. Here n=20; Which is [N/2th observation + (N/2 +1)th observation]/2 observation [ Arranging in ascending order ] = 4+5/2 = 4.5 iii.  Mode = Most of the reparative observation = 0 B. i) Range = Maximum – Minimum = 23 – 0 = 23
Image of page 1

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

View Full Document Right Arrow Icon
ii) Standard deviation = 6.3185 C. i.   P15 = 15 * 20 / 100 = 15 * 1/5 = 3 rd Point, Mean of 3 rd , 4 th Point Which is 0.5 ii.   P66 = 66 * 20 / 100 = 66 / 5 = 13.2 = 13 Which is 8 iii. Minimum = 0, Q1 = 1 , Q2 = 4.5, Q3 = 9 , Maximum = 23 Q2 A. i.  ∑ f= 75 ∑ fx = 5847 Mean = ∑ fx / ∑ f = 77.96 ii.  33rd data point falls in value 86, Hence median is 86 iii. The data point which has maximum value is the mode point. Here frequency value 15 is the maximum & Data point is 87 B.
Image of page 2
i.  Find Standard deviation ii.   Q1 = (N+1)/4 = 75+1/4 = 19 th point = 84 , Q3 = 3(N+1)/4 = 19* 4 = 76 th point = 92 IQR = Q3 − Q1 = 92-84 = 8 iii.
Image of page 3

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

View Full Document Right Arrow Icon
Image of page 4
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