Question

On average 20% of the gadgets produced by a factory are mildly defective. I buy a box of 100 gadgets. Assume this is a random sample from the production of the factory. Let A be the event that less than 15 gadgets in the random sample of 100 are mildly defective. (a) Give an exact expression for p(a), without attempting to evaluate it. (b) Use either the normal or the Poisson approximation, whichever is appropriate, to give an approximation of p(a).

Answer 1

Answer:

a) $P(A)=P(x$

b) $P(A)\approx 0.085$

Step-by-step explanation:

The proportion of defective gadgets for this factory is p=0.2.

If a sample of n=100 gadgets is taken from the production, we can model the amount of defectives gadgets in this sample as a binomial random variable.

The probability of having k defectives in the sample can be written as:

$P(x=k)=\binom{100}{k}p^k(1-p)^{100-k}$

Then, we can write the expression to calculate the probabilities of A: "less than 15 gadgets are mildly defective":

$P(A)=P(x$

If we approximate this binomial distribtution to a normal distribution, we can calculate the new parameters as:

$\mu=np=100*0.2=20\\\\\sigma=\sqrt{np(1-p)}=\sqrt{100*0.2*0.8}=\sqrt{16}=4$

The continuity factor is applied for the change from a discrete distribution (binomial) to a continous distribution (normal). Then we have:

$P(A)=P(X$

Now, we can calculate the probability P(A):

$z=(X-\mu)/\sigma=(14.5-20)/4=-5.5/4=-1.375\\\\P(A)=P(X$