solutions hw 4

# solutions hw 4 - 1c 2c 3c 4b 5b 6d 7d 8a 9c The mean of the...

This preview shows page 1. Sign up to view the full content.

1 c The mean of the population data is μ = ∑x ∕ N The variance of the population data is Item x A 33 5.76 B 45 92.16 C 54 345.96 D 27 70.56 E 18 302.76 177 817.2 μ = 35.4 163.44 2 c The mean of the sample data is The variance of the sample data Item x A 33 1 B 45 169 E 18 196 96 ∑(x − x‾)² = 366 32 183 3 c The mean of the means equals the mean: 35.4 N = 5, n = 3 The variance of the means is Since the population is finite, you must include the Finite Population Correction Factor in the formula FPCF = (N − n) ∕ (N − 1 N = 5 n = 3 FPCF = 0.5 163.44 27.24 4 b C(N, n) = C(36, 6) = 1,947,792 5 b 200 The mean of the means equals the mean: 200 6 d 50 5 7 d μ = 80 σ = 30 Here we are dealing with the distribution of the population data. P(x < 77) z = (x - μ) ∕ σ z = -0.1 P(z < -0.1) = 0.4602 8 a P(74 < x < 86) Again, we are dealing with the population data. x = 86 x = 74 z = 0.2 z = -0.2 P(-0.2 < z < 0.2) = 0.1585 9 c 30.8 129.2 10 b μ = 80 σ = 30 n = 25 6 z = -0.5 (z < -0.5) = 0.3085 11 d μ = 80 σ = 30 n = 64 P(72.5 < x‾ < 87.5) 3.75 72.5 z = -2 87.5 z = 2 P(-2 < z < 2) = 0.9545 12 c μ = 80 σ = 30 n = 64 Find the 90% MOE. MOE = (1.64)(3.75) = 6.15 73.85 86.15 13 c μ = 100 σ = 16 n = 50 2.263 98 102 z = -0.88 z = 0.88 P(-0.88 < z < 0.88) = 0.6211 14 a μ = 100 σ = 16 n = 100 1.6 98 102 z = -1.25 z = 1.25 P(-1.25 < z < 1.25) = 0.7887 15 c μ = 2.20 σ = 0.10 n = 50 0.014 2.18 2.22 z = -1.41
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• 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.

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

• 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.

Dana University of Pennsylvania ‘17, Course Hero Intern

• 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.

Jill Tulane University ‘16, Course Hero Intern