The mean and the standard deviation of the sampled population are, respectively, 77.4 and 4.0. find the mean and standard deviation of the sampling distribution of the sample mean x overbarx for samples of size n equals 225n=225. round your answers to one decimal place.

μₓ = 77.4

σₓ = 4.0

Step-by-step explanation:

We have to find the mean and standard deviation of the sampling distribution of the sample mean x, given mean and standard deviation of sampled population:

Mean=μ= 77.4

Standard deviation = σ = 4.0

and sample size = 255 n= 255 => n= 255/255 => n=1

so, the mean of sample mean x will be same as mean of the sampled population:

μₓ = μ = 77.4

and the standard deviation of sample mean x will be standard deviation of the sampled population divided by square root of sample size:

σₓ = σ / √ n

  = 4.0 / √1

  = 4.0

a complex number is a number that can be expressed in the form a + bi, where a and b are real numbers, and i is a solution of the equation x2 = −1. because no real number satisfies this equation, i is called an imaginary number. has no real solution, since the square of a real number cannot be negative.