MATH SOLVE

8 months ago

Q:
# Five percent of all vcrs manufactured by a large electronics company are defective. a quality control inspector randomly inspects groups of 10 vcrs from the production line. using the tables for the binomial probability distribution, what is the probability that exactly one of the vcrs is defective in any one group of 10 vcrs?

Accepted Solution

A:

Binomial

Binomial distribution can be used because the situation satisfies all the following conditions:1. Number of trials is known and remains constant (n=10)2. Each trial is Bernoulli (i.e. exactly two possible outcomes) (defective/normal)3. Probability is known and remains constant throughout the trials (p=5%)4. All trials are random and independent of the others (assumed from context)The number of successes, x, is then given by[tex]P(x)=C(n,x)p^x(1-p)^{n-x}[/tex]where[tex]C(n,x)=\frac{n!}{x!(n-x)!}[/tex]

Substituting values, p=0.05, n=10, X=exactly 1

for X=1 (defective out of n)

P(X=1)=C(10,1)0.05^1*(1-0.05)^(10-1)

=10!/(1!9!)*0.05*0.95^9

=10*0.05*0.0630249

=0.315125 (to 6 places of decimal)

Binomial distribution can be used because the situation satisfies all the following conditions:1. Number of trials is known and remains constant (n=10)2. Each trial is Bernoulli (i.e. exactly two possible outcomes) (defective/normal)3. Probability is known and remains constant throughout the trials (p=5%)4. All trials are random and independent of the others (assumed from context)The number of successes, x, is then given by[tex]P(x)=C(n,x)p^x(1-p)^{n-x}[/tex]where[tex]C(n,x)=\frac{n!}{x!(n-x)!}[/tex]

Substituting values, p=0.05, n=10, X=exactly 1

for X=1 (defective out of n)

P(X=1)=C(10,1)0.05^1*(1-0.05)^(10-1)

=10!/(1!9!)*0.05*0.95^9

=10*0.05*0.0630249

=0.315125 (to 6 places of decimal)