{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

exam 2 review solutions

# exam 2 review solutions - Review Two 1 Blood type is...

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

Review Two 1) Blood type is inherited. If both parents carry genes for the O and A blood types, each child has a 25% chance of getting two O genes, and so having blood type O. Different children inherit independently of each other. Using an observation of 60 children whose parents are carries, a) Find μ = n*p = 60*0.25 = 15 and σ = 3541 . 3 75 . * 25 . * 60 * * = = q p n . b) Find P(X = 15) = binomdist(15, 60, 0.25, false) = 0.118228 = 11.82% P(X > 45) = 1 - P(X 45) = 1 - binomdist(44, 60, 0.25, true) = 0 c) Create a histogram of this data. 0 0.02 0.04 0.06 0.08 0.1 0.12 0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 40 42 44 46 48 50 52 54 56 58 60 Histogram binomial normal d) Find P(X = 15) and P(x > 45) using a normal approximation. P(X = 15) = normdist(15, 60*0.25, sqrt(60*0.25*0.75), false) = 0.118941 = 11.89% P(X > 45) = 1 - P(X 45) = 1 - normdist(45, 60*0.25, sqrt(60*0.25*0.75), true) = 0 e) Create a histogram of the normal approximation. Answer is in b) f) Check for normality using > 5, histogram, and Emperical Rule. (1) Histograms are a bell curve (2) One 18.3541 70.3825 % > 68% 11.6459 Two 21.7082 0.948984 >93.75% 8.29179 6 Three 25.0623 1 0.998291 >99.26% 4.93769 4

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

View Full Document
2) For f X (x) = + otherwise x for x x 0 40 12 81 26 2 and using Integrating.xls I. Show that f X (x) could be a p.d.f. for some continuous random variable. FUNCTION 0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0 10 20 30 40 50 x f ( x ) Formula =x/(26*sqrt(81+x^2)) The graph is between 0 and 1 and the sum equals One. II. Find E(x). = 26.365623 Formula =x^2/(26*sqrt(81+x^2)) Integration interval a = 12 and b = 40 III. Find P(22 x < 32) = 36.430045% Integration interval a = 22 and b = 32 IV. Find P(x < 15) = 9.587906% Integration interval a = 12 and b = 15 V. Find V(x) = 64.187248 =(x – 26.365623)^2*(x/(26*sqrt(81+x^2)) 3) Let Y be a normal random variable with μ Y = 5.77 and σ Y = 0.15. Create changeable random samples of size 400. Arrange the Excel sheet so that it automatically computes the mean and standard deviation of each sample. = NORMINV(RAND(), 5.77, 0.15) then drag ten across and 40 down Use mean = average(cells) and standard deviation = stdev(cells) to compute automatically. 4) Use Integrating.xls to find a number z 3 such that P ( Z z 3 ) = 0.9900. z 3 = NORMSINV(0.99) = 2.326348
5) In problem #4, you found a number z 3 such that P ( Z z 3 ) = 0.9900. Use this to modify our computations and find a 98% confidence interval for the mean of county administrators’ salaries. Use the sample that our administrator took to compute the 98% confidence interval. Starting point: You want to find a positive number b such that P ( - b μ X + b ) = 0.98. (You do not need to reproduce all of the steps in the computation, simply indicate the confidence interval and the possibly different outcome that you find at the 90% level.) P ( x - b μ X x + b ) = 0.98 where n n z n z b σ σ σ = = = 326348 . 2 / 3 3 . thus P ( n x σ - 326348 . 2 μ X n x σ + 326348 . 2 ) = 0.98 6) Use your results from problem #5 to modify the sheet Intervals in Salary.xls so that it computes the end points of 98% confidence intervals for the mean of X .

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 17

exam 2 review solutions - Review Two 1 Blood type is...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online