Question

A) in the pic

B) if the probability of a success is 0.35, the probability of fewer than 4 successes is what?

C) if the probility of success is 0.15, the standard deviation of the random variable is what?

Score: 0 of 4 pts 3 of 6 (4 complete) 5.2.28-T Assuming the binomial distribution applies with a sample size of n=11, find the values below. a. the probability of 5 or more successes if the probability of a success is 0.80 b. the probability of fewer than 4 successes if the probability of a success is 0.35 c. the expected value of the random variable if the probability of success is 0.15 d. the standard deviation of the random variable if the probability of success is 0.15 a. If the probability of a success is 0.80, the probability of 5 or more successes is (Round to four decimal places as needed.)

Answer 1

Solution - Lif it is oo and Binomial Distribution Dismbuchon - let X is a discrete random said to have Ginomial distribution with two P then it's pmf becomes, Yamable parameters p(x=x) = a pu (1-2) 1-4 12011,2,-.in osPS) XNB(0,P) He know that Expected value = E(21) = np and Standard deviation = 6 - VOPCI-P) @ Here sample size (n) = 11 and poobability of a success (P) = 0.80 XNB (,0) = XNB( 11 0.80) Then to find Pls or more success ) = P(x>s) je. P( x75) = 1-P(XLS) - TP(x=0) + P(X=1) + P(X=2) + P(x=3) + P(x=4) / (") 60,80) ° (1-0.80) "-0 + ( (0.80)' (1-0.80) (180) (1-0,80) "-34 +(960.80)? (1-0,80) "-2 + (!) (0.80) 4 (1-0,80) 11-4 ] = 1 0.0000000 2048 + 0.0000090112 + 0.0000\80224 + 0.000216268 + 0.00173015 ] - 0.0002433 = 0.99975 - P(x>s) = 0.9997

6 sample papbability size (0) 11 and of success (P) = 0,25 T ic. XUB (nIP) = XNB( 11, 0.35) To find P fewer than 4 successes) = ? lie. P( x 4 ) = P(x=0) + P(x=1) + P(x=2) + P(x=3) = (0) (0.35)° (1-0.35) "-0 + (1) (0.35)' (1-0,35) "-) - 11-3 + ( ) (0.35)2 (1-0.35) "-2 + ( 60.35) 3 (1-0.35) = 0.0087507 + 0,051831 + 0.139546 + 0,22562) PCX <U) 04255 Sample size in all and probability of success (P) = 0115 ie XB (NIP) - XNB(0, 0.15) To find expected value. - Expected value = E(0) = np = (11) (0.15) = 1.65 e Sample size (n) = 1 and probability of success (P) = 0115 ie XOB (NIP) = XNB(11,0.15) Then, to find standard deviation. - Standard deviation = 6 2 MP (1-P) = V (II) (0.15) (1-0015) 1.4025 1.1843 1. Standard deviation =