Question

In a shipment of 20,000 toys (called robot chickens) 600 of the toys are defective. Suppose that 20 toys are selected at random (without replacement) for inspection, and let X denote the number of defective toys found.

a) The distribution of the random variable X is (choose one)

i) Binomial

ii) hypergeometric

iii) Poisson

iv) Normal

v) Exponential

vi) Uniform

b) Find P(X≤6).

c) Which distribution from those listed in part (a) can be used as an approximation to the distribution of X if we assume they are selected with replacement? With this approximation find P(X≤6).

d) If we make the assumption that they are selected with replacement as in part (c), since 600/20,000=0.03 is small (hint - rare) and since we have select enough components (hint – large n) the distribution in (c) can be approximated by which distribution listed in (a) ? With this approximation find P(X≤6)

Answer 1

a) ii) hypergeometric (since sampling is done without replacenent)

b)

from excel: P(X<=6) =HYPGEOM.DIST(6,600,20,20000,TRUE) =0.9999988

c)

here this can be approximated with binomial with parameter n=20 and p=600/20000=0.03

P(X<=6)=

∑x=0a     (nCx)px(1−p)(n-x)    =

0.9999988

if using ti-84 use command :binomcdf(20,0.03,6)

if using excel use command :binomdist(6,20,0.03,true)

d)

Poisson distribution with parameter λ =np=20*0.03 =0.6

P(X<=6)=

∑x=0x   {e-λ*λx/x!}=

0.9999967

if using ti-84 use commannd :poisson(0.6,6)

if using excel use commannd :poisson(6,0.6,true)