Discover answers to your questions with Westonci.ca, the leading Q&A platform that connects you with knowledgeable experts. Join our platform to get reliable answers to your questions from a knowledgeable community of experts. Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently.

A random variable X has a mean of 120 and a standard deviation of 15 a random variable Y has a mean of 100 and a standard deviation of 9. if c and y are independent approximately what is the standard deviation of X plus Y

Sagot :

Answer:

The standard deviation is 17.5

Step-by-step explanation:

Let S be the sum of X and Y

S = X + Y

So we want to calculate the standard deviation of S

The standard deviation of S will be the square root of the sum of the variances of X and Y

Kindly recall that variance is the square of standard deviation

The variance of X will be 15^2 = 225

The variance of Y will be 9^2 = 81

So the standard deviation of the sum will be

√(81 + 225)

= √(306)

= 17.5

The standard deviation of X + Y is 17.5

The given parameters are:

[tex]\bar x =120[/tex] --- the mean of X

[tex]\sigma_x =15[/tex] -- the standard deviation of X

[tex]\bar y = 100[/tex] --- the mean of Y

[tex]\sigma_y = 9[/tex] --- the standard deviation of Y

The standard deviation of X + Y is then calculated as:

[tex]\sigma_{x +y} = \sqrt{\sigma_x^2 + \sigma_y^2}[/tex]

This gives

[tex]\sigma_{x +y} = \sqrt{15^2 + 9^2}[/tex]

Evaluate the exponents

[tex]\sigma_{x +y} = \sqrt{225 + 81}[/tex]

[tex]\sigma_{x +y} = \sqrt{306}[/tex]

Take the square root of 306

[tex]\sigma_{x +y} = 17.5[/tex]

Hence, the standard deviation of X + Y is 17.5

Read more about standard deviation at:

https://brainly.com/question/475676

Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. We appreciate your time. Please come back anytime for the latest information and answers to your questions. Get the answers you need at Westonci.ca. Stay informed with our latest expert advice.