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;
![P(X=r) = \binom{n}{r}p^{r} (1-p)^{n-r} ; x = 0,1,2,3,.....](https://tex.z-dn.net/?f=P%28X%3Dr%29%20%3D%20%5Cbinom%7Bn%7D%7Br%7Dp%5E%7Br%7D%20%281-p%29%5E%7Bn-r%7D%20%3B%20x%20%3D%200%2C1%2C2%2C3%2C.....)
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%
<em>LET X = Number of bulbs that are defective</em>
So, it means X ~ Binom(
)
Now, probability that between 5 and 7 (both inclusive) bulbs from the sample are defective is given by = P(5
X
7)
P(5
X
7) = P(X = 5) + P(X = 6) + P(X = 7)
= ![\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}](https://tex.z-dn.net/?f=%5Cbinom%7B16%7D%7B5%7D%5Ctimes%200.20%5E%7B5%7D%20%5Ctimes%20%281-0.20%29%5E%7B16-5%7D%2B%20%20%5Cbinom%7B16%7D%7B6%7D%5Ctimes%200.20%5E%7B6%7D%20%5Ctimes%20%281-0.20%29%5E%7B16-6%7D%2B%20%5Cbinom%7B16%7D%7B7%7D%5Ctimes%200.20%5E%7B7%7D%20%5Ctimes%20%281-0.20%29%5E%7B16-7%7D)
= ![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}](https://tex.z-dn.net/?f=4368%20%5Ctimes%200.20%5E%7B5%7D%20%5Ctimes%200.80%5E%7B11%7D%20%2B8008%20%5Ctimes%200.20%5E%7B6%7D%20%5Ctimes%200.80%5E%7B10%7D%20%2B11440%20%5Ctimes%200.20%5E%7B7%7D%20%5Ctimes%200.80%5E%7B9%7D)
= 0.1948
Hence, the probability that between 5 and 7 (both inclusive) bulbs from the sample are defective is 0.1948.