Question

In: Statistics and Probability

On any given day, Otis Campbell has a 31% chance of walking into the mayberry jail...

On any given day, Otis Campbell has a 31% chance of walking into the mayberry jail uninvited. Suppose we randomly choose 15 days to observe if Otis walks into the jail or not.

a) Check that this is a binomial probability. There is four checkpoints to meet

b) What is the probability that Otis shows up exactly 3 days? Round to three decimals.

c) What is the probaility that Otis shows up one or more days? Round to three decimals.

d) u=7.75 and o=2.31. Find the usual range. Write as an interval.

e) is it usual for Otis to show up to the Mayberry jail 10 days out of 15, uninvited? Why?

Expert Solution

a) As there is an equal probability of walking into the mayberry jail uninvited for all 15 days, therefore the random variable process here is a sum of n independent and identical events. Therefore this is a case of a binomial distribution given as:

b) The probability that Otis shows up exactly 3 days is computed using the binomial probability function here as:

Therefore 0.1579 is the required probability here.

c) The probability that Otis shows up one or more days is computed here as:

= 1 - Probability that Otis dont show up on any day

= 1 - (1 - 0.31)¹⁵ = 0.9962

Therefore 0.9962 is the required probability here.

d) The usual range for a variable lies within 2 standard deviations of the mean.

Mean - 2*Std Dev = 7.75 - 2*2.31 = 3.13
Mean + 2*Std Dev = 7.75 + 2*2.31 = 12.37

therefore the usual range here is given as: (3.13, 12.37)

e) The probability of having a value 10 or more is computed here as:

As the probability here is 0.0048 < 0.05, therefore this is an unusual event here.

orchestra answered 2 years ago

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