Respuesta :
Answer:
(a) P(X [tex]\leq[/tex] 20) = 0.9319
(b) Expected number of defective light bulbs = 15
(c) Standard deviation of defective light bulbs = 3.67
Step-by-step explanation:
We are given that a manufacturer of Christmas light bulbs knows that 10% of these bulbs are defective. It is known that light bulbs are defective independently. A box of 150 bulbs is selected at random.
Firstly, the above situation can be represented through binomial distribution, i.e.;
[tex]P(X=r) = \binom{n}{r} p^{r} (1-p)^{2} ;x=0,1,2,3,....[/tex]
where, n = number of samples taken = 150
r = number of success
p = probability of success which in our question is % of bulbs that
are defective, i.e. 10%
Now, we can't calculate the required probability using binomial distribution because here n is very large(n > 30), so we will convert this distribution into normal distribution using continuity correction.
So, Let X = No. of defective bulbs in a box
Mean of X, [tex]\mu[/tex] = [tex]n \times p[/tex] = [tex]150 \times 0.10[/tex] = 15
Standard deviation of X, [tex]\sigma[/tex] = [tex]\sqrt{np(1-p)}[/tex] = [tex]\sqrt{150 \times 0.10 \times (1-0.10)}[/tex] = 3.7
So, X ~ N([tex]\mu = 15, \sigma^{2} = 3.7^{2})[/tex]
Now, the z score probability distribution is given by;
Z = [tex]\frac{X-\mu}{\sigma}[/tex] ~ N(0,1)
(a) Probability that this box will contain at most 20 defective light bulbs is given by = P(X [tex]\leq[/tex] 20) = P(X < 20.5) ---- using continuity correction
P(X < 20.5) = P( [tex]\frac{X-\mu}{\sigma}[/tex] < [tex]\frac{20.5-15}{3.7}[/tex] ) = P(Z < 1.49) = 0.9319
(b) Expected number of defective light bulbs found in such boxes, on average is given by = E(X) = [tex]n \times p[/tex] = [tex]150 \times 0.10[/tex] = 15.
Standard deviation of defective light bulbs is given by = S.D. = [tex]\sqrt{np(1-p)}[/tex] = [tex]\sqrt{150 \times 0.10 \times (1-0.10)}[/tex] = 3.67
