Question

6. A box contains 3 quarters and 7 nickels. Suppose two coins are randomly selected without replacement from this box. (a) (3 points) Complete the the probability distribution table below for the total amount in cents. Use reduced fraction for probabilities. T. total amounts in cents P(T) (b) (2 points) Graph the probability distribution histogram. 30 50 60 70€ 5¢ (c) (2 points) Find 10¢ . (d) (2 points) Find o. (e) (2 points) Find the exact value for o'. (e) () (2 points) How much are you willing to pay to play this game? Explain. Page 4 of 4 Study Guide 17 Total Points: 50

Answer 1

6. A box contains 3 quarters and The number of ways we can 7 nickels. select a two coins lo! 10! n(s) = 10cz-Lo 21 (10-82)! 2!8) ..115)= 45 ways. let T be the total amount is cents. The value of T are locent, 30 cent, 50 cents. T=10 if we select both nickel. number of ways we can select two nicked by (two nickel) = 72 = 71 = 42 = 21 215! [PCT= 10) = 2 1/5 = 7/15 one T= 30 cents if we select one quarter and nicked. ::hci quarter finickel) = 3€, * 7c, = 387 2 21 :: PCT=30)= 2175 T = 50; if we select both quarter, an(Both quarter) = 3 = 3 1 1 - 3 45

ap(T2501-21 135 =ts q) . The probability distribution of T is T 110¢ 30¢ 50¢ PIT) |7/15 7/15 Yıs distribution b) graph of probability a histogram is 0507 .. 0.40 0.307 (Lod a 20 0.ro 0(0,0) 10 50 Karis. 30 Amount incent (T) c) The expected ralue of Tis MECKT) = {to P (T=t) = 10x778 + 30x7/5 + 50x'ts = 7 8 + 2,00 + 50 is

d) The standard deviation is 6= Juae(t) where vae(t) = xt_p(724) - 42 = (1 ofx g +(301 x + 650jas- (20.67)2 = 100** + 900* + 2- 427.2489 t 6300 2500 - 700 T5 -427.2489 I a -427.22 = 9500-427.2489 = 633,33-42712489 = 206.0811 :: 6 - 1206.08.11 [6=1436] Pdl) 62= varct) 1 20810811 f) The expected value is u = 20.67 Therefore, we are willing to pay 20.67~ 21 cent to play this game.