Question

10. A class has 8 students who are to be assigned seating by lot. What is the probability that the students will be arranged in order from shortest to tallest?

11. Minimum sample size

12. Probability?

Points: 5 10) A class has 8 students who are to be assigned seating by lot. What is the probability that the students will be arranged in order from shortest to tallest? (Assume that no two students are the same height.) Show your work below. D) 0.1000 A) 0.00019841 C) 0.0000248 B) 0.00024802 Use the given data to find the minimum sample size required to estimate the population proportion. Points: 5 11) Margin of error: 0.025; confidence level: 99%; P and unknown minimum sample size Show your work. Estimate the indicated probability. Points: 5 12) A product is manufactured in batches of 120 and the overall rate of defects is 5%. Estimate the probability that a randomly selected batch contains less than 6 defects. probability Explain how you found the answer.

Answer 1

Question -10

Answers 8 to cam students There are 8! ways arrange in any order + Only ܥܐ te way arrange student from shortest to tallest Probability 12 - 100 8x8x6x5x4x3x2x) - 0.0000248 40320 Option (c) is correct

Question -11

Ahvena -0.025 2 Margin of Enor ef Pin is unknown we We take Å=0.5 At 99% confidence level "Z critica" value is x= 1-0.99 = 0.01 a 4/2 = 0.011 -0.005 不 Talr - Z 0.005 = 2.576 (from z -table) Sample size (n) is.

Sample Size (n) is 3 Lala 2 Cc: -1-P) n = * (P) *(1-1) E (2536) + (0.9)+ (0,5) (103.04) *(0,25) 10617.2416 *10.25) = 2654.3104 n 2655 (rounded up to nearest integer)

Question -12

Eliven 4 e- n. 120 pe 50% = 0.05 np 120 (0.05) = 6 > 5 n (1-P) - 12000-95) 120(0,95) = 114 Her Binomial variable "X" ix approximately normally disturbated. Mean CH) - пP - 120 (0-05) –6 SD (o) = Vn-(p.(1-P V120.00-09).C0-45) = 157 15.7 - 2.38.15

5 - P(x<6) = p(x < 5.5) Cusing continuity Conection) 5.5-6 - P ( X < 33.3845 = P(Z <=056) = P(za - 0.21) = 0.4168338 = 0.4168 (rounded to 4 decimals)

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