Question

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1. 120 points! Four machines produce the total output of a factory, Machine 1 produces 30%, machine 2 produces 25% machine 3 produces 12% and machine 4 produces 13% of the output. 5% of the output of machine lis defective, 8% from machine 2 is not defective, 3% from machine 3 is defective and 4% from machine 4 is not defective. If a finished item is selected at rindom, a. What is the probability of it being nor defective h. Given that the item is defective what is the probability that it was produced by Machine 4 2. 20 Points Let X denotes the amount of space occupied by an article placed in a loft packing container. The pdf of X is kx(1 - x) otherwise Find the constant b. Construct the cdf of X € Calculate the expected value of X 3. [10 Points] Consider an exam with 28 multiple choice questions. Assume that each question has three solution choices but only one of them is correct. A student who didn't study for the exam is going to make a guess. 1. What is the probability that he guesses only 15 questions correctly? What is the probability that he guesses at least 15 questions correctly c. What is the probability that he guesses more than 10 and less than 15 questions correctly? d How many questions do you expect him to guess incorrectly!

Answer 1

And (1) For i=1,2,3,4, let Midlenote the event that item is produced by machin i and Dd be the event produced item is defective. Given that PLM) = 0.30, PCM250,25, P (M)=0.124 PCM4)=0.33 , PODIM, 1-0.05, P(DIM2770.92 , PEDIM3170.03, PCDIMA) = 0.96 If an item is randomly selected. Man Can The prob of it being not defective ( D ) P(51= plonmi) where PC Önmi) = PODIMO pomi) So Plonmi I = P(DIMIR(MI) = 0.9580.30 = 0,2850 PLOOM2) = PC / M27 PLM 2) = 0.00 x0.25= 0,0200 PCDOM)= PCÕIM31PCM3 ) = 0,97X0.12 = 0.1164 Pc Doma) = pubima) pema) = 204 x 0.33= 0.013 2

- (2) 0 mi k ro " Hence PD = 0,2850 + 0.0200 + 0.1164 + 0.0132 = 0,4346 (6) Given that item is defective, the prob that it was produced by machine 4 is PCM41D) PCDOM4) PIDIM4). P(M4) PD) IPC) tolaut 4 wor] -0.96X 0.33 0.3168 075604 0.5603 day -0.4346 0.5654 The pdf of X il f(x) = 1 k 3c061–0 ; 0x<) otherwise cap Since fer is valid pdf, 10 that So'fende = 1 So k x8(1-x)dt = 8 dx *f*9dx] = K[1099 - 1997 Ek Gb) So that k=go (6) The tedf of 900 & tolgood word nila F(x) = Pr (X EX 1 = k t° LO (!-tldt of choice kl 19 - LIST = * [ 30 TÓLobo Memory old water te Flxis go x9 Fore) = 29 10-9x) > 0SX4) W (c) the expected value of x is lebe-out an E(X)=xflvdx = 1' xf(x) dx = k [ 29c1-x) de ou parang *[ 1%-15", 12 kl to the promo to 90 110 go kx9/ So - KX 9 T