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Probability and Statistics
Name__________________________

Spring 1998 Flex-Mode and Flex-Time 45-733

Practice Final

Keith Poole

(5 Points)

1. The proportion of people that have a certain type of cancer in a large
population is known to be .002. If we take a random sample of 2000 people,
what is the probability that exactly 5 people have the type of cancer.

Poisson approximation: **l = np** = 2000*.002 = 4

**P(X = 5) = F(5) - F(4)** = .785 - .629 = .156

Probability and Statistics
Name__________________________

Spring 1998 Flex-Mode and Flex-Time 45-733

Practice Final

Keith Poole

(10 Points)

2. Suppose we have the continuous probability distribution:

Find the maximum likelihood estimator for b.{ (b^{4}/6)x^{3}e^{-bx}f(x) = { x > 0 { 0 otherwise

Hence

_{^}_ b = 4/X_{n}

Probability and Statistics
Name__________________________

Spring 1998 Flex-Mode and Flex-Time 45-733

Practice Final

Keith Poole

(10 Points)

3. Assume thatFindE(X_{1}) = 3 VAR(X_{1}) = 6 COV(X_{1}, X_{2}) = -2 E(X_{2}) = 4 VAR(X_{2}) = 8 COV(X_{1}, X_{3}) = 0 E(X_{3}) = 5 VAR(X_{3}) = 8 COV(X_{2}, X_{3}) = 1

4*6 + 25*8 + 1*8 - 20*(-2) + 4*0 - 10*1 = 262

Spring 1998 Flex-Mode and Flex-Time 45-733

Practice Final

Keith Poole

(10 Points)

4. We wish to estimate the average height of women first year GSIA MBA students. Suppose the population standard deviation is known to be 3.5 inches. We take a random sample of 36 women. What is the probability that the difference between the sample mean and the true mean will not exceed .4 inches._ We Want P(|X_{n}- m| £ .4) = P(-.4 £ X_{n}- m £ .4) = P[(-.4)/(3.5/6) £ (X_{n}- m)/(3.5/6) £ (.4)/(3.5/6)] = P(-.69 £ Z £ .69) = F(.69) - F(-.69) = 1 - F(-.69) - F(-.69) = 1 - .2451 - .2451 = .5098

Spring 1998 Flex-Mode and Flex-Time 45-733

Practice Final

Keith Poole

(10 Points)

5. The checkout times for customers at the Squirrel Hill Giant Eagle are
independent random variables with a mean of 3.5 minutes and a standard deviation
of 1.5 minutes. What is the approximate probability that it will take more
than 3 hours to process the orders of 50 people.

**E(X _{i}) = 3.5, VAR(X_{i}) = 1.5^{2}**

Hence,

P(Z > .47) = 1 - F(.47) = F(-.47)

Spring 1998 Flex-Mode and Flex-Time 45-733

Practice Final

Keith Poole

(10 Points)

6. Suppose we have the bivariate discrete probability distribution:

{ c(3x + y^{2}) { x = 1,2,3 f(x,y) = { y = 1,2,3 { { 0 otherwise

- Find c
y 1 2 3 ------------ 1 | 4 7 12 | 23 | | x 2 | 7 10 15 | 32 | | 3 |10 13 18 | 41 | | --------------- 21 30 45 | 96 Hence

**c = 1/96**

- Find
**r(X, Y)**

**E(X)**= (1/96)*(1*23 + 2*32 + 3*41) = 210/96

**E(Y)**= (1/96)*(1*23 + 2*30 + 3*45) = 216/96

**E(X**= (1/96)*(1*23 + 4*32 + 9*41) = 520/96^{2})

**E(Y**= (1/96)*(1*23 + 4*32 + 9*41) = 546/96^{2})

**E(XY)**= (1/96)(1*1*4 + 1*2*7 + 1*3*12 + 2*1*7 + 2*2*10 + 2*3*15 + 3*1*10 + 3*2*13 + 3*3*18)= 468/96

**r(X, Y)**= [(468/96) - (210/96)(216/96)]/{[(520/96) - (210/96)^{2}] [(546/96) - (216/96)^{2}]}^{1/2}

Spring 1998 Flex-Mode and Flex-Time 45-733

Practice Final

Keith Poole

(10 Points)

7. Suppose we read in ** Robber Barrons** that 85 percent of 200
first year Flex-Time and Flex-Mode students would rather have 10 root canals
than retake 45-733. Compute 95 percent confidence limits for the true proportion of first
years who would rather have the root canals.

Hence, .85 ± 1.96[.85*.15/200]= .85, n = 200,_{^}pa= .05,Z= 1.96_{.025}

or (.8005, .8995)

Spring 1998 Flex-Mode and Flex-Time 45-733

Practice Final

Keith Poole

(10 Points)

8. Two new drugs were given to patients with ulcers. The first drug lowered the acid concentration of 15 patients by an average of 25 points with a standard deviation of 8 points. The second drug lowered the acid concentration of 14 patients by 19 points with a standard deviation of 7 points. Determine 95 percent confidence limits for the difference in the mean reductions in acid concentration. Assume the measurements are normally distributed with equal variances.

Hence,_ X_{n}= 25 s_{x}= 8 n = 15 _ a = .05, t_{.025, 27}= 2.052 Y_{m}= 19 s_{y}= 7 m = 14 s^{2}= (14*64 + 13*49)/(15 + 14 - 2) = 56.778 s = 7.535

Spring 1998 Flex-Mode and Flex-Time 45-733

Practice Final

Keith Poole

(10 Points)

9. We take a random sample of potholes in Pittsburgh city streets and
measure the depth of each pothole. The measurements in centimeters are:

135 101 99 201 121 89 199 148 91Assume that the depth of Pittsburgh potholes is normally distributed. Find 99 percent confidence limits for the true depth of the population of Pittsburgh potholes.

Hence, 131.566 ± 3.355*(43.609/3) = 131.566 ± 48.796_ X_{n}= 131.556, s = 43.609, a = .01, t_{.005, 8}= 3.355

Or (82.77, 180.362)

Spring 1998 Flex-Mode and Flex-Time 45-733

Practice Final

Keith Poole

(10 Points)

10. Suppose the time that it takes to complete a Prob-Stat Final Exam is normally distributed. We draw a random sample of 14 first year Flex-Time/Flex-Mode students and find that their exam times (in minutes) are:

155 145 100 98 76 178 159 161 102 132 144 146 179 140Construct 99 percent confidence limits for population variance of the exam times.

Hence,

and the limits are: (13*991.720/29.8194 , 13*991.720/3.56503) = (432.348 , 3616.340)

Spring 1998 Flex-Mode and Flex-Time 45-733

Practice Final

Keith Poole

(10 Points)

11. Suppose we are drawing a random sample of 36 from a normal distribution. Given the hypothesis test:

IfH_{0}: m = 175 s^{2}= 2500 H_{1}: m = 195

Now, since_ a = P(X_{n}> m_{0}+ c | m = m_{0}) _ = P[(X_{n}- 175)/(50/6) > (175 + c - 175)/(50/6)] = = P(Z > 6c/50) = 1 - F(6c/50) = .01

Hence, 2.33 = 6c/50 and

_ b = P(X_{n}< m_{0}+ c | m = m_{1}) _ = P[(X_{n}- 195)/(50/6) < (175 + 19.417 - 195)/(50/6)] = = P(Z < -.07) = F(-.07) = .4721

Spring 1998 Flex-Mode and Flex-Time 45-733

Practice Final

Keith Poole

(10 Points)

12. We draw a random sample of size 30 from a normal distribution and we
find that **s ^{2}** = 143.5. Compute 95 and 99 percent confidence
limits for

For

so that the limits are (91.017 , 259.33)

For

so that the limits are (79.516 , 317.161)

Spring 1998 Flex-Mode and Flex-Time 45-733

Practice Final

Keith Poole

(10 Points)

13. We give identical Prob-Stat I final exams to first year and second year MSIA students. We draw a random sample of 9 first years and find that their exam scores are:111 121 131 99 123 125 102 130 119We draw a random sample of 8 second years and find that their exam scores are:

127 90 88 132 130 101 97 131Assume that the exam scores are normally distributed in both populations. Test the null hypothesis that the variances of the two populations are the same against the alternative hypothesis that the variances are not the same (

First years:

Test statistic:

Since .221 < .342 < 4.90,

****************************

or, Test statistic:

Since 1/4.90 = .204 < 2.920 < 4.53,

Spring 1998 Flex-Mode and Flex-Time 45-733

Practice Final

Keith Poole

(10 Points)

14. We wish to draw a random sample from a normal distribution with mean_ P[|X_{n}- m| £ .5] ³ .99

_ P[-.5 £ X_{n}- m £ .5] = _ P[-.5/(6/n^{1/2}) £ (X_{n}- m)/(6/n^{1/2}) £ .5/(6/n^{1/2})] = P[-.5/(6/n^{1/2}) £ Z £ .5/(6/n^{1/2})] = F(.5/(6/n^{1/2}) - F(-.5/(6/n^{1/2}) = 2F(.5/(6/n^{1/2}) - 1 ³ .99 F(.5/(6/n^{1/2}) ³ .995 Now, since F(2.58) = .9951, therefore, .5/(6/n^{1/2}) ³ 2.58, hence n ³ [(2.58*6)/.5]^{2}= 958.52 and n = 959