Seventy percent of rivets from vendor A meet a certain strength specification, and 80% of rivets from vendor B meet the same specification. If 510 rivets are purchased from each vendor, what is the probability that more than 775 of the rivets meet the specifications?

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:

Check the conditions for both the population as follows:

Thus, the Normal approximation to Binomial can be used.

Since both sample are independent then the distribution of X₁ + X₂ is:

Compute the probability of the event (X₁ + X₂ > 775) as follows:

*Use a standard normal table.

Thus, the probability that more than 775 of the rivets meet the specifications is 0.2327.

hey there!

if she's saving for six months, let's divide the total number we need for the trip by six, dividing it into the amount that needs to be saved for each month.

500 ÷ 6 =

she needs to save at least this amount each month, so we should use this symbol: ≥

let x be the amount she saves each month.

your final answer is x ≥

hope this !