exam 2 review solutions

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

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Unformatted text preview: 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 = n*p = 60*0.25 = 15 and = 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. 0.02 0.04 0.06 0.08 0.1 0.12 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 2) For f X (x) = + otherwise x for x x 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 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 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 Y = 5.77 and 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 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 X x + b ) = 0.98 where n n z n z b = = = 326348...
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exam 2 review solutions - Review Two 1) Blood type is...

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