Answered

Discover new perspectives and gain insights with IDNLearn.com. Join our interactive community and get comprehensive, reliable answers to all your questions.

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

MrRoyalidn

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

Thank you for participating in our discussion. We value every contribution. Keep sharing knowledge and helping others find the answers they need. Let's create a dynamic and informative learning environment together. Thank you for choosing IDNLearn.com. We’re committed to providing accurate answers, so visit us again soon.