Probability related to Normal Distribution

In summary, the problem involves two normally distributed variables, A and B, with given means and variances. The goal is to find the probability of B being greater than A. The solution involves finding the probability of a new variable Y, which is the difference between B and A, being greater than 0. Using the standard normal distribution, the probability is calculated to be 0.9568.
  • #1
songoku
2,319
331

Homework Statement


Variable A has mean 55 and variance 9, variable B has mean 65 and variance 25. If A and B are normally distributed, find P (B > A)


Homework Equations


z = (x - μ) / σ



The Attempt at a Solution


Can this be solved? What is the meaning of P (B > A)? The probability of random variable B is bigger than random variable A?

Thanks
 
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  • #2
I suppose the problem assumes A and B are statistically independent. Then the formal way to do this is:
[itex]P(B>A) = \int_{b=-\infty}^{+\infty} \int_{a=-\infty}^{b}f_A(a)f_B(b)dadb[/itex]
This is a very nasty integral.
 
  • #3
songoku said:

Homework Statement


Variable A has mean 55 and variance 9, variable B has mean 65 and variance 25. If A and B are normally distributed, find P (B > A)


Homework Equations


z = (x - μ) / σ



The Attempt at a Solution


Can this be solved? What is the meaning of P (B > A)? The probability of random variable B is bigger than random variable A?

Thanks

Yes, P(B > A) means exactly what you said. It can be solved quite easily if A and B are independent, but I cannot give more details now: first you have to make a serious attempt to solve it yourself.
 
  • #4
hikaru1221 said:
I suppose the problem assumes A and B are statistically independent. Then the formal way to do this is:
[itex]P(B>A) = \int_{b=-\infty}^{+\infty} \int_{a=-\infty}^{b}f_A(a)f_B(b)dadb[/itex]
This is a very nasty integral.

Sorry I don't get the idea how to do the integration and why can it be like that

Ray Vickson said:
Yes, P(B > A) means exactly what you said. It can be solved quite easily if A and B are independent, but I cannot give more details now: first you have to make a serious attempt to solve it yourself.

I don't have idea how to start. The question I saw until now always contained number such as P (B > 23). Please give me clue how to start solving this problem. What is the first step I need to take? I can't find the value of z because I don't know the value of the random variable.

Thanks
 
  • #5
songoku said:
Sorry I don't get the idea how to do the integration and why can it be like that



I don't have idea how to start. The question I saw until now always contained number such as P (B > 23). Please give me clue how to start solving this problem. What is the first step I need to take? I can't find the value of z because I don't know the value of the random variable.

Thanks

Look at the random variable Y = B - A.
 
  • #6
Ray Vickson said:
Look at the random variable Y = B - A.

OK maybe I get it

Let: Y = B - A

E(Y) = E(B) - E(A) = 10
σ(Y) = σ(B) + σ(A) = 34

P(B > A) = P(B - A > 0) = P(Y > 0) = P(Z > -1.715) = 0.9568

Do I get it right?

Thanks
 
  • #7
songoku said:
OK maybe I get it

Let: Y = B - A

E(Y) = E(B) - E(A) = 10
σ(Y) = σ(B) + σ(A) = 34

P(B > A) = P(B - A > 0) = P(Y > 0) = P(Z > -1.715) = 0.9568

Do I get it right?

Thanks

It looks right.
 
  • #8
Ray Vickson said:
It looks right.

ok thanks
 

Related to Probability related to Normal Distribution

What is the Normal Distribution?

The Normal Distribution is a probability distribution that is commonly used in statistics to represent a pattern of data. It is also known as the Gaussian Distribution or the Bell Curve due to its characteristic shape.

What are the characteristics of the Normal Distribution?

The Normal Distribution is symmetric, meaning that the data is evenly distributed on both sides of the mean. It also follows the 68-95-99.7 rule, where approximately 68% of the data falls within one standard deviation of the mean, 95% within two standard deviations, and 99.7% within three standard deviations.

How is the Normal Distribution related to Probability?

The Normal Distribution is used to calculate probabilities for continuous random variables. It allows us to determine the likelihood of a particular value or range of values occurring within a dataset.

What is the Central Limit Theorem and how is it related to the Normal Distribution?

The Central Limit Theorem states that the sum of a large number of independent and identically distributed random variables will tend towards a Normal Distribution, regardless of the underlying distribution of the variables. This makes the Normal Distribution a useful tool for analyzing data from a wide range of sources.

How is the Normal Distribution used in real-world applications?

The Normal Distribution is used in many real-world applications, including quality control, financial analysis, and risk assessment. It is also commonly used in social sciences and natural sciences to analyze data and make predictions. Additionally, it is used in medical research to determine the effectiveness of treatments and identify potential health risks.

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