Question

1. MINITAB QUESTION. Let X be a hypergeometric random variable with N= 50, M= 18, and n = 30. (a) Cumulative probabilities are those of the form P(X 5x). Use MINITAB to print the cumulative probability table for X by following the instructions in Box A. Include the session window output with your assignment (b) Use the output from part (a) to calculate each of the following probabilities. Explain your reasoning. (1) P(X S9) (ii) P(X<9) (ii) P(X 2 7) (iv) P(85 X 511) Box A Click Calc Make Patterned Data - Simple Set of Numbers. In the "Store patterned data in" box type C1. In the "From first value" box type 0. In the "To last value" box type 18. Click OK. Click Calc Probability Distributions Hypergeometric. Enter the "Population size", "Event count in population", and "Sample size" in the appropriate boxes. In the "Input column:" box type C1. Click OK.

Answer 1

(a)Box A:

step 1:

Minitab - Untitled X O File Edit Data Calc Stat Graph Editor Tools Window Help Assistant ..... DlhA+Y W TOOOM Simple Set of Numbers Session Store patterned data in: C1 X From first value: 0 To last value: 18 Welcome to In steps of: 1 Number of times to list each value: 1 TIT Number of times to list the sequence: 1 Worksheet 1 Select + C1 C8 Help OK Cancel 1 1 N 3 4 u Current Worksheet: Worksheet 1 Editable

Step 2:

x wl Minitab - Untitled X File Edit Data Calc Stat Graph Editor Tools Window Help Assistant DE | *%&)2 . h TOO Hypergeometric Distribution х C1 Session Probability Cumulative probability Inverse cumulative probability Population size (N): 50 Welcome to 18 Event count in population (M): Sample size (n): 30 Input column: c1/ Optional storage: Worksheet 1 Input constant: + C1 Select C8 Optional storage: 0 1 Help OK Cancel 2 NA 3 4 3 5 4 Current Worksheet: Worksheet 1 Editable

Obtained probabilities:

Welcome to Minitab, press F1 for help. Cumulative Distribution Function Hypergeometric with N = 50, M = 18, and n = 30 0 1 2 5 6 8 P ( XS x) 0.00000 0.00000 0.00000 0.00000 0.00006 0.00067 0.00482 0.02376 0.08366 0.21676 0.42640 0.66097 0.84666 0.94951 0.98854 0.99833 0.99986 1.00000 1.00000 10 11 12 13 14 15 16 17 18

(b) :P(x<q) = 0.21676 (from the obtained table, X=9) ii. P(x<9) = P(X + 8) => PCX<9)= 0.08366 Cfrom the table, x=8)....Answer, . P(x> +) = 1-P(x<+) |-PCX 26) -1-0.00482 (from the table, x=6 x=6) -0.99518 ....... iv. P(8 < X<11) = P(X=11) – P(x<8) P(X<11) - P(x<+) From table, x= 11 gives P(x411) = 0.66094 2-4 gires P(x + 4) =0.023 46 Hence, P(x411) -- P(X4) 0.66097 -0.02376 0. 63791 - Answers Answer -