contestada

The distribution of the amount of money in savings accounts for florida residents has an average of 2,200 dollars and a standard deviation of 900 dollars. suppose that we take a random sample of 100 florida residents and ask them how much they have in their savings account. the sampling distribution of the sample mean amount of money in a savings account is

Respuesta :

syed514
The sampling distribution of the sample mean amount of money in a savings account is :
approximately Normal, with a mean of 2200 and a standard error of 900 because  they don't give you enough information in the problem. The problem states the Sx of 900 dollars and the mean to be 2200 dollars. They don't give you the random sample number.

Standard deviation shows the dispersion of data. The sampling distribution of the sample's mean amount of money in a savings account is $90.

What is the standard deviation?

The standard deviation shows us how varying the data is from the mean.

We know that the residents have an average of $2,200 while their standard deviation is $900. Now, in order to calculate the sample mean amount of money in a savings account for 100 random residents, we will find the standard error of mean which is given by the formula,

[tex]\sigma_x = \dfrac{\sigma}{\sqrt{n}}[/tex]

Substitute the values,

[tex]\sigma_x = \dfrac{900}{\sqrt{100}}\\\\\sigma_x = 90[/tex]

Hence, the sampling distribution of the sample mean amount of money in a savings account is $90.

Learn more about Standard Deviation:

https://brainly.com/question/12402189

Otras preguntas