Question

A national standard requires that public bridges over feet in length must be inspected and rated every 2 years. The rating scale ranges from 0​ (poorest rating) to 9​ (highest rating). A group of engineers used a probabilistic model to forecast the inspection ratings of all major bridges in a city. For the year​ 2020, the engineers forecast that ​% of all major bridges in that city will have ratings of 4 or below. Complete parts a and b. a. Use the forecast to find the probability that in a random sample of major bridges in the​ city, at least 3 will have an inspection rating of 4 or below in 2020. ​P(x​3) nothing ​(Round to five decimal places as​ needed.)

Answer 1

Answer:

The answer is below

Step-by-step explanation:

A national standard requires that public bridges over 20 feet in length must be inspected and rated every 2 years. The rating scale ranges from 0​ (poorest rating) to 9​ (highest rating). A group of engineers used a probabilistic model to forecast the inspection ratings of all major bridges in a city. For the year​ 2020, the engineers forecast that 4​%of all major bridges in that city will have ratings of 4 or below.

a. Use the forecast to find the probability that in a random sample of major bridges in the​ city, at least 3 will have an inspection rating of 4 or below in 2020.

Answer:

This problem is a probability binomial distribution and it can be solved using the formula:

$P(X=x)=C(n,x)p^xq^{n-x}\\\\q=1-p,C(n,x)=\frac{n!}{x!(n-x)!}$

Hence the solution to the problem is given as:

P(x ≥ 3) = 1 - P(x < 3) = 1 - [ P(x=0) + P(x=1) + P(x = 2)]

Given that p = 4% = 0.04, q = 1 - p = 1 - 0.04 = 0.96, n = 10. Hence:

$P(x=0)=C(10,0)*0.04^{0}*(0.96)^{10-0}=0.6648\\\\P(x=1)=C(10,1)*0.04^{1}*(0.96)^{10-1}=0.277\\\\P(x=2)=C(10,2)*0.04^{2}*(0.96)^{10-2}=0.0519\\\\P(x\geq 3)=1-[0.6648+0.277+0.0519]=0.0063$