Question

Jacco Macacco was a monkey that used to fight dogs in London 1820 years ago.   Assuming he a 65% probability of winning his matches, answer the following:

a) what are the chances he could win 14 matches in a row before his first loss?

b) winning exactly 10 out of his first 14 matches (assuming he did not die in the ones he lost)

c)  say that recorded 10,000 monkey vs dog fights. What is the probability that between 6,000 and 7,000 of the fights had monkey winners?

d) say each monkey weighted in at 10lbs with a standard deviation of 2lbs. If you weighed 100 monkey’s, what would the standard deviation of the total be? What about the standard deviation of the average? How many monkey’s would you need to weight to have a 68% chance of the average weight be between 9.5lbs and 10.5lbs?

Answer 1

a)

The probability of winning is

P(win) = 0.65

The probability of lossing the match is

P(loose) = 1 - P(win) = 1 - 0.65 = 0.35

The chances he could win 14 matches in a row before his first loss is

b)

Here we need to use binomial distribtuion with parameters n=14 and p=0.65.

The probability that he is winning exactly 10 out of his first 14 matches is

c)

Here X has binomial distribution with following parameters,

n=10000 and p=0.65

Since sample size is very large so we need to use normal approximation. Using normal approximation, the mean of X is

The standard deviation

The z-score for X = 6000-0.5 = 5999.5 is

The z-score for X = 7000+0.5 = 7000.5 is

The required probability is

d)

Let W shows the weight of monkey. Here we have

Let T shows the total weight. The standard deviation of total weight is

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Let n shows the sample size. The standard deviation of average weight is

According to normal distribution, 68% data values lie within one standard deviation of mean. Now