Question

1. The following is the probability distribution of the number of customers buying bicycle accessories at a particular shop in one day. 1 4 X P(X = x) 0 0.05 2 0.15 3 0.25 5 0.30 6 0.05 0.10 a (a) (b) (c) Compute the value of a. Find the probability that the shop did not sell any bicycle accessory in one day. Find the probability that at least three customers bought from the shop in one day. Find the probability that at most five customers bought from the shop in one day. Find the expected value of X. Find the variance and standard deviation of X. (d) (e) 2. A machine produces parts of which 8% are defective. If a random sample of 10 parts produced by this machine contains two or more defectives, the machine is dow for repairs. Find the probability that the machine will be shut down based on this sampling plan. 3. The number of students coming in complaining of stomach problems during a typical day at the Student Health Center is Poisson distributed with mean u. If the probability that no student comes in complaining of stomach problems in one day is 0.015. (a) (b) Calculate u. Find the probability that more than 6 students come in complaining of stomach problems in 2 days.

Answer 1

P(0) +P(1) +PC2) +-P() = 1 → 0.05+0:10 +0)5+0.25+ a +0:30+ 0.05=1 => a=0.10 b) P(X=0) = 0.05 (C) PCX >3) = P(X=3) +P (X=4) + f(x=5) +P(X=6) = 0.70 d) PCX 55) = 1-p(x=6) = -0.05 = 0.95 e) E(X)= EXP() = 0 PO+LPOO +2.P(2)+.. EX)= x 3:35 (6) Varta = =(x-8) P(x) 0-3-35) PCO + (1-3-35) P(1) + 28.8575 Standard Deviation = 5288575 28.8575 -5372 2. Machine will be shutdown if x2 P(x>2) |- P(o)-P (1) = 1- C(10,0) (008)°(0.92) - C (10,1) (0 8'(0.93) = (-0.4344 -0.3777 = 0.1879 3. @ P(x=k) where a is mean. LK P (x=0) = 0.015 > 2 = 4.2 (b) Mean number of students in a days = 8.4 P(x>6) = 1- E P(x=k) = 1-0.1572 = 0.8427 10 K = K=0