Respuesta :
Answer:
Probability that between 5 and 7 (both inclusive) bulbs from the sample are defective is 0.1948.
Step-by-step explanation:
We are given that quality control inspector has drawn a sample of 16 light bulbs from a recent production lot.
Suppose 20% of the bulbs in the lot are defective.
The above situation can be represented through Binomial distribution;
[tex]P(X=r) = \binom{n}{r}p^{r} (1-p)^{n-r} ; x = 0,1,2,3,.....[/tex]
where, n = number of trials (samples) taken = 16 light bulbs
r = number of success = between 5 and 7 (both inclusive)
p = probability of success which in our question is % of bulbs in
the lot that are defective, i.e; 20%
LET X = Number of bulbs that are defective
So, it means X ~ Binom([tex]n=16,p=0.20[/tex])
Now, probability that between 5 and 7 (both inclusive) bulbs from the sample are defective is given by = P(5 [tex]\leq[/tex] X [tex]\leq[/tex] 7)
P(5 [tex]\leq[/tex] X [tex]\leq[/tex] 7) = P(X = 5) + P(X = 6) + P(X = 7)
= [tex]\binom{16}{5}\times 0.20^{5} \times (1-0.20)^{16-5}+ \binom{16}{6}\times 0.20^{6} \times (1-0.20)^{16-6}+ \binom{16}{7}\times 0.20^{7} \times (1-0.20)^{16-7}[/tex]
= [tex]4368 \times 0.20^{5} \times 0.80^{11} +8008 \times 0.20^{6} \times 0.80^{10} +11440 \times 0.20^{7} \times 0.80^{9}[/tex]
= 0.1948
Hence, the probability that between 5 and 7 (both inclusive) bulbs from the sample are defective is 0.1948.
