Respuesta :
Answer:
Probability that more than 25 chips are defective is 0.1075.
Step-by-step explanation:
We are given that the manufacturing of semiconductor chips produces 2% defective chips. Assume the chips are independent and that a lot contains 1000 chips.
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 = 1000 chips
r = number of success = more than 25
p = probability of success which in our question is % of
defective chips, i.e; 2%
LET X = Number of chips that are defective
SO, X ~ Binom(n = 1000, p = 0.02)
Now, here we can't find the probability that more than 25 chips are defective using binomial distribution because the sample size is very large here (n > 30), so we will use Normal approximation to find the respective probability.
So, mean of binomial distribution = E(X) = [tex]n\times p[/tex]
= [tex]1000 \times 0.02[/tex] = 20
Standard deviation of binomial distribution = S.D.(X) = [tex]\sqrt{n\times p \times (1-p)}[/tex]
= [tex]\sqrt{1000 \times 0.02 \times 0.98 }[/tex]
= 4.43
Let Y = Number of chips that are defective for normal approximation;
So, Y ~ Normal([tex]\mu=20,\sigma = 4.43[/tex])
The z-score probability distribution for normal distribution is given by;
Z = [tex]\frac{Y-\mu}{\sigma}[/tex] ~ N(0,1)
Now, probability that more than 25 chips are defective is given by = P(Y > 25) = P(Y > 25.5) --------------{using continuity correction}
P(Y > 25.5) = P( [tex]\frac{Y-\mu}{\sigma}[/tex] > [tex]\frac{25.5-20}{4.43}[/tex] ) = P(Z > 1.24) = 1 - P(Z [tex]\leq[/tex] 1.24)
= 1 - 0.89255 = 0.1075
The above probability is calculated using z table by looking value of x = 1.24 in z-table.
Therefore, the probability that more than 25 chips are defective is 0.1075.
