Statistics Data Manipulation

1. DATA MANIPULATION AND PLOTTING

2. PROBABILITY

Question

1. Let X be the number of years the manager makes a correct prediction. Then let n be the total amount of years the fund manager is used consecutively.

The probability the fund manager predicts correctly whether the exchange will rise of fall over a given year is p=0.55. (Given)

P=0.55 n=3

a) PrX=3 When n=3

PrX=3=n!X!n-X! PX (1-P)n-X

i) Plugging in the probabilities and variables given.

= 3!3!3-3! 0.553 (1-0.55)3-3

= 660! 0.16638 0.450

ii) So the probability of successfully predicting 3 years in succession.

PrX=3=0.16638 =16.6%

b) PrX>2=PrX=3orPrX=2

i) First need to calculate pr(X=2)

PX=2=n!X!n-X! PX (1-P)n-X

ii) Plugging in the probabilities and variables given.

= 3!2!3-2! 0.552 (1-0.55)3-2

= 621! 0.3025 0.451

iii) So the probability of successfully predicting 2 out of the 3 years.

PrX=2= 0.408375 =41%

iv) Using the calculated probability from question a and Pr(X=2)

PrX≥2=PrX=3+Pr⁡(X=2)

v) Plugging in the probabilities and variables given.

= 0.16638+(0.408375)

vi) So the probability of successfully predicting at least 2 out of the next 3 years correctly.

PrX≥2=0.574755=57%

c) PrX=0 When n=3

PrX=0=n!X!n-X! PX (1-P)n-X

i) Plugging in the probabilities and variables given.

= 3!0!3-0! 0.550 (1-0.55)3-0

= 613! 1 0.453

ii) So the probability of successfully predicting 0 years out of the next 3.

PrX=0=0.091125=9%

d) The fund manager is not worth his upfront annual fee of 5%.

2. Let X be the number of server failures in a giver year. And the mean number of failures in a year is µ=λ=1.3

a) Prx=1When λ=1.3

Prx=1= e-λλXX!

i) Plugging in the probabilities and variables given....