Question 1085660: A manufacturer of batteries knows that 10% of the batteries produced by a particular
production line are defective. A sample of 200 batteries is removed for testing.
a) What is the probability that exactly 20 batteries in the sample are defective?
b) What is the probability that at least 20 batteries in the sample are defective?
Answer by mathmate(429)   (Show Source):
You can put this solution on YOUR website! Question:
A manufacturer of batteries knows that 10% of the batteries produced by a particular
production line are defective. A sample of 200 batteries is removed for testing.
a) What is the probability that exactly 20 batteries in the sample are defective?
b) What is the probability that at least 20 batteries in the sample are defective?
Solution:
We know from the question that it is a binomial distribution because the following conditions are met:
1. known and constant probability throughout trials (p=0.1)
2. known number of trials (200).
3. Bernoulli trial (exactly two possible outcomes).
4. Trials are independent of each other.
The probability of exact x defectives out of sample size N is
P(x)=C(N,x)(p^x)(1-p)^(N-x)
(a) for 20 defectives out of 200, x=20, N=200, p=0.1, C(200,20)=200!/(20!180!)=1613587787967350073386147640
So
P(20)=1613587787967350073386147640*0.1^20*0.9^180
=0.0936
(b) for at least 20 defectives, we subtract from 1 the sum of probabilities from 0 to 19 defectives.
Thus
P(>=20)=1 - sum(P(i)) for i=0 to 19
=1-
(7.05508*10^-10
+1.56780*10^-8
+1.73328*10^-7
+1.27108*10^-6
+6.95561*10^-6
+3.02955*10^-5
+1.09401*10^-4
+3.36884*10^-4
+9.03037*10^-4
+0.00214
+0.00454
+0.00872
+0.01526
+0.02452
+0.03638
+0.05013
+0.06440
+0.07745
+0.08749
+0.09312)
=1-0.46554
=0.53446
However, this is tedious to do the above calculations. Most of the time, you would look up a binomial distribution table, or use a statistical package such as R which you can download for free,
https://www.r-project.org/
or use an intermediate level calculator such as the TI83.
On the other hand, if you have learned about normal approximation, both parts (a) and (b) can be approximated using normal approximation, with continuity corrections. One of the conditions of normal approximations is N>40.
We start with calculating the equivalent
mean=np=200*0.1=20
variance=npq=200(0.1)(1-0.1)=200*.09=18
standard deviation=sqrt(variance)=sqrt(18)=4.246
(a)
P(20)
=Z((20.5-20)/4.246)-Z((19.5-20)/4.246)
=0.546907-0.453093
=0.09381
which is not very far from 0.0936 (exact answer)
(b)
P(>=20)
=1-P(<20)
=1-Z(19.5)
=1-.453093
=0.54691
still close to 0.53446 above.
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