Question

3. Suppose that X has a Poisson distribution with mean μ=15. Use the 'cdf' command MTB > cdf; SUBC 〉 poisson 15. Find PX 〈 10)-[3] and P(15 X 20)= [3] 4. Suppose that Y has a hypergeometric distribution with parameters N = 20, M = 6, and n = 4, Use the command MTB > cdf 3; hypergeometric 20 6 4. a. [3] to find P(Y 53) = b. 3) Use the similar command to find P(Y > 2) =

Answer 1

The step-by-step instructions to obtain Poisson distributi on probabilities and cumulative probabilities using Minitab18 Store the data from 1to 20 in the column C1 name the column isX in the Minitab workbook Go to Calc ->Probability distributions >Poisson *Check the Probability K Check the cumulative probability * Enter the value of Mean 15 Input column with the vaniable named X. *Storage column with the pd (P(X-x)once and cdf (P(Xs). the next time you give command. xClick ok. Poisson Distribution output using Minitab18

C1 сг C3 0 0.000000 0.000000 1 0.000005 0.000005 2 0.000034 0.000039 3 0.000172 0.000211 4 0.000645 0.000857 5 0.001936 0.002792 6 0.004839 0.007632 7 0.010370 0.018002 8 0.019444 0.037446 9 0.032407 0.069854 10 0.048611 0.118464 11 0.066287 0.184752 12 0.082859 0.267611 13 0.095607 0.363218 14 0.102436 0.465654 15 0.102436 0.568090 16 0.096034 0.664123 17 0.084736 0.748859 18 0.070613 0.819472 19 0.055747 0.875219 20 0.041810 0.917029 7 10 12 13 14 15 16 17 18 19 20 21

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Let X be the poisson random vari able. The cumative distribution function is x! The mean of the Poisson distributi on is u-15 Letl be the random variable followesa Poisson distribution is The probability that the random variable _X is less than 10 is e-3(15)e-5/151 e15 (15 2! 0+0+0+0.0002 +0.0006+0.00 19 + 0.0048+0.0104+ 0! 9! 0.0194 +0.0324 Since, on using the values of the poisson di stribution table using Minitab18 0.0699 P(X <10) 0.0699 Therefore, the poisson di stribution random variable is less than 10 i:s P(X <10)=10.0699]

C1 сг C3 0 0.000000 0.000000 1 0.000005 0.000005 2 0.000034 0.000039 3 0.000172 0.000211 4 0.000645 0.000857 5 0.001936 0.002792 6 0.004839 0.007632 7 0.010370 0.018002 8 0.019444 0.037446 9 0.032407 0.069854 10 0.048611 0.118464 11 0.066287 0.184752 12 0.082859 0.267611 13 0.095607 0.363218 14 0.102436 0.465654 15 0.102436 0.568090 16 0.096034 0.664123 17 0.084736 0.748859 18 0.070613 0.819472 19 0.055747 0.875219 20 0.041810 0.917029 7 10 12 13 14 15 16 17 18 19 20 21

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Let X be the poisson random variable The cumative di stribution function is The mean of the Poisson distribution is μ-15. Let Y be the ran dom vari able follows a Poisson distribution is The probability that the random variable X is less than or equal to 15 and greater than or equal to 20 is Since, on using the values of the 0.917029-0.56809 po isson distribution table using the output of Minitab 18 in the part (a) 0.34894 P(15s Xs20) 0.3484 Therefore, the probability that the random variable X is less than or equal to 15 and greater than or equal to 20 is P(15SX S20)-03484