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95-796 HOMEWORK 1

Name:NIZEYIMANA Jean Paul

Carnegie Mellon University in Rwanda

Andrew ID:jnizeyim

Course: Statistics for IT Managers

1. a)

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Looking at the Histogram of volume, we can see that the distribution is positively skewed. Looking at the

Box plot of volume, we can see that there is an outlier (Row 2: volume = 31415.9).

b) Descriptive Statistics: Volume

Variable N N* Mean SE Mean StDev Minimum Q1 Median Q3 Maximum

Volume 30 0 11735 1340 7340 1616 7245 9027 16811 31416

- Mean = 11735

- Median = 9027

- Standard deviation = 7340

- Variance = the square of the standard deviation = 73402 = 53875600

- Range = maximum minimum = 31416 1616 = 29800

- Interquartile range = Q3 Q1 = 16811 7245 = 9566

Based on these statistics, the hypothesis stated that the distribution is positively skewed is confirmed

because the mean > median (11735 > 9027).

c) Diameters of first six trees: 7, 11, 12, 16, 18, 20

- Mean = (7+11+12+16+18+20) / 6 = 84 /6 = 14

- Median = (12+16) /2 = 14

- Standard deviation:

Deviations: -7, -3, -2, 2, 4, 6

Squared deviations: 49, 9, 4, 4, 16, 36

Variance: S2 = (49+9+4+4+16+36) / (6-1) = 118/5 = 23.6

Standard deviation: S = 6.23 = 4.86

- Range = 20-7 = 13

2. Conditional probability and Bayes rule

We are given the following probabilities:

P(use drugs) = 0.03 this means that P(dont use drugs) = 0.97

P(positive | uses drugs) = 0.95 this means that P(negative | uses drugs) = 0.05

P(positive | doesnt use drugs) = 0.2 this means that P(negative | doesnt use drugs) = 0.8

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Using Bayes rule:

a) P(uses drugs | positive) =

drugs)usetP(don'drugs)tdoesn'|P(positivedrugs)P(usedrugs)uses|P(positive

drugs)P(usedrugs)uses|P(positive

+

=97.0*2.003.0*95.0

03.0*95.0

+

= 0.0285 / 0.2225 = 0.128090

b) P(uses drugs | negative) =

drugs)usetP(don'drugs)tdoesn'|P(negativedrugs)P(usedrugs)uses|P(negative

drugs)P(usedrugs)uses|P(negative

+

=97.0*8.003.0*05.0

03.0*05.0

+

= 0.0015 / 0.7775 = 0.001929

c) P(uses drug | A and B are positive) = P(uses drug | A is positive) * P(uses drug | B is positive)

=drugs)usetP(don'drugs)tdoesn'|positiveisP(Adrugs)P(usedrugs)uses|positiveisP(A

drugs)P(usedrugs)uses|positiveisP(A

+*

drugs)usetP(don'drugs)tdoesn'|positiveisP(Bdrugs)P(usedrugs)uses|positiveisP(B

drugs)P(usedrugs)uses|positiveisP(B

+

=97.0*2.003.0*95.0

03.0*95.0

+

*97.0*2.003.0*95.0

03.0*95.0

+

= 0.128090* 0.128090= 0.0164070481

d) P(B is positive | A is positive) = P(B is positive | use drugs) P(use drugs | A is positive) + P(B is

positive | dont use drug) P (dont use drug | A is positive)

=

negative)isP(Bnegative)isB|drugsusetP(don'positive)isP(Bpositive)isB|drugsusetdon'P(

positive)isP(Bpositive)isB|drugsusetdon'P(

+

* = 0.128090+

negative)isP(Bnegative)isB|drugsP(usespositive)isP(Bpositive)isB|drugsusesP(

positive)isP(Bpositive)isB|drugsusesP(

+*

0.87191

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=81.0*8.019.0*87191.0

19.0*87191.0

+

* 0.128090 +81.0*05.019.0*128090.0

19.0*128090.0

+

*0.87191

= 0.2036 * 0.128090 + 0.3754 * 0.87191 = 0.3534

3. Discrete and continuous random variables

a)Manager 1:

We have the interval [$5,000, $35,000]

Mean: = ($5,000 + $ 35,000) / 2 = $ 20,000

Standard deviation = ($35,000 - $ 5,000) / 12 = $8660.2540

Manager 2:

The mean and the standard deviation of his groups monthly expenses are the same as those estimated.

Mean: = $21,000

Standard deviation: =$2,000

Manager 3:

Mean: = )(xpx

= 0.2($15,000) + 0.4($20,000) + 0.3($25,000) + 0.1($30,000) = $21500

Variance: 2 = )()(2

xpx

= ((-6500)2 0.2 + (-1500)2 0.4 + (3500)2 0.3 + (8500)2 0.1)) = $20250000

Standard deviation: = 2 = 20250000 = $4500

b) Using Minitab, we can calculate these probabilities.

The steps are:

Select Calc > Probability Distributions > NormalSelect Cumulative Probability

Fill-in the values in the dialog box (mean, standard deviation and the input constant)

ClickOK

Manager 1:

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i) From Minitab,

Cumulative Distribution Function

Normal with mean = 20000 and standard deviation = 8660.25

x P( X

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22000 0.691462 (less than $22,000)

Thus the probability that is greater than $22,000 will be 1- 0.691462 = 0.308538

iii)Cumulative Distribution Function

Normal with mean = 21000 and standard deviation = 2000

x P( X

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Fill-in the values in the dialog box (mean, standard deviation and the input constant)

ClickOK

The input constant will be 20% = 0.2

For manager 1:

Inverse Cumulative Distribution Function

Normal with mean = 20000 and standard deviation = 8660.25

P( X

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Inverse Cumulative Distribution Function

Normal with mean = 21000 and standard deviation = 2000

P( X

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