(a). Suppose X has a Poisson distribution with parameter =11.44. Find the follow

Question

(a). Suppose X has a Poisson distribution with parameter =11.44. Find the following.

(i). The mean of X. (2 points).

(ii).The standard deviation of X.

(b).The number of lightning strikes in a year at the top of particular mountain has a poison distribution with parameter =2.9.Find the following.

(i). The probability that in a randomly selected year the number of striking is less than or equal to 5.

(ii). The probability that in a randomly selected year the number of striking is greater than 5.

(c).The probability that a call received by a certain switchboard will be a wrong number is 1 per 100 calls. Use poison approximation to the binomial distribution to find the probability that among 120 calls received by the switchboard,

(i). There is no wrong numbers.

(ii). There is exactly one wrong number.

(iii). There are exactly two wrong numbers.

(iv). There are exactly three wrong numbers.

(v). There are at most three wrong numbers.

show work please

Explanation / Answer

(a). Suppose X has a Poisson distribution with parameter =11.44. Find the following.
(i). The mean of X. (2 points).

Note that for Poisson variables,

mean = = 11.44 [ANSWER]

(ii).The standard deviation of X.

For Poisson variables,

standard deviation = sqrt() = sqrt(11.44) = 3.382306905 [ANSWER]

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(b).The number of lightning strikes in a year at the top of particular mountain has a poison distribution with parameter =2.9.Find the following.

(i). The probability that in a randomly selected year the number of striking is less than or equal to 5.

Using a cumulative poisson distribution table or technology, matching          
          
u = the mean number of successes =    2.9      
          
x = the maximum number of successes =    5      
          
Then the cumulative probability is          
          
P(at most   5   ) =    0.925826198 [ANSWER]

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(ii). The probability that in a randomly selected year the number of striking is greater than 5.

Note that P(more than x) = 1 - P(at most x).          
          
Using a cumulative poisson distribution table or technology, matching          
          
u = the mean number of successes =    2.9      
          
x = our critical value of successes =    5      
          
Then the cumulative probability of P(at most x) from a table/technology is          
          
P(at most   5   ) =    0.925826198
          
Thus, the probability of at least   6   successes is  
          
P(more than   5   ) =    0.074173802 [ANSWER]

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