Respuesta :
Answer:
The probability that more than 775 of the rivets meet the specifications is 0.2327.
Step-by-step explanation:
Let X₁ and X₂ be the number of rivets that meet specification from vendor A and B respectively.
The proportions of rivets that meet specification from vendor A is, p₁.
The proportions of rivets that meet specification from vendor B is, p₂.
The sample of rivets selected from each vendor is, n₁ = n₂ = 510.
The random variables X₁ and X₂ follow Binomial distribution.
But, since the sample selected from each population is large a Normal approximation to Binomial can be used if:
- np ≥ 10
- n(1 - p) ≥ 10
Check the conditions for both the population as follows:
[tex]n_{1} p_{1}=510\times 0.70=357>10\\n_{1} (1-p_{1})=510\times (1-0.70)=153>10\\n_{2} p_{2}=510\times 0.80=408>10\\n_{2} (1-p_{2})=510\times (1-0.80)=102>10\\[/tex]
Thus, the Normal approximation to Binomial can be used.
[tex]X_{1}\sim N(n_{1}p_{1},\ n_{1}p_{1}(1-p_{1}))\\X_{2}\sim N(n_{2}p_{2},\ n_{2}p_{2}(1-p_{2}))[/tex]
Since both sample are independent then the distribution of X₁ + X₂ is:
[tex]X_{1}+X_{2}\sim N(n_{1}p_{1}+n_{2}p_{2},\ n_{1}p_{1}(1-p_{1})+n_{2}p_{2}(1-p_{2}))\\X_{1}+X_{2}\sim N(357+408,\ 107.1+81.6)\\X_{1}+X_{2}\sim N(765,\ 188.7)[/tex]
Compute the probability of the event (X₁ + X₂ > 775) as follows:
[tex]P(X_{1}+X_{2}>775)=P(\frac{(X_{1}+X_{2})-\mu}{\sigma}>\frac{775-765}{\sqrt{188.7}})\\=P(Z>0.73)\\=1-P(Z<0.73)\\=1-0.7673\\=0.2327[/tex]
*Use a standard normal table.
Thus, the probability that more than 775 of the rivets meet the specifications is 0.2327.
