(1 point) Consider the probablility model with sample space {A,B,C) and P(A)=0.3, P(B)=0.3, P(C)=0.4. Then (a) P(B or C) =
(b) P(A and B ) =

Solved

Show answers

Ответ:

Kiki18love

04.03.2021

Mathematics

0.2420

Step-by-step explanation:

Let x be a random variable that represents the number of cars passed through the intersection per day .

As per given , we have

$\mu=550000$

$\sigma=100,000$

We assume that the number of cars on the interchange is approximately normally distributed.

∵ $z=\dfrac{x-\mu}{\sigma}$

Then for x= 620,000, $z=\dfrac{620000-550000}{100000}=0.7$

The probability more than 620,000 use the interchange on a random day :-

$P(x620000)=P(z0.7)=1-P(z\leq0.7)\\\\=1-0.7580363=0.2419637\approx0.2420$

Hence, the probability more than 620,000 use the interchange on a random day= 0.2420

0,0(0 оценок)

Ask your question to the AI advisor

Ai-bot is an expert in any field and is the perfect companion for reliable and useful answers and advice on a variety of topics, including science, history, technology, art, sports, health, culture and more.

Ask your question

More tips

Answers on questions: Mathematics

Ask an AI advisor a question

I want advice

(1 point) Consider the probablility model with sample space {A,B,C) and P(A)=0.3, P(B)=0.3, P(C)=0.4. Then (a) P(B or C) = (b) P(A and B ) =