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.

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

dgsistemasp0pffb dgsistemasp0pffb · Answer 2 · 2019-10-25T15:51:14+02:00

Answer:

μₓ = 77.4

σₓ = 0.3

Step-by-step explanation:

The sample mean is an estimator of the population mean, the proportion observed in the sample is an estimator of the proportion in the population.

So, the sample mean is the same as the population.

However, the standard deviation of the sample is achieved by the equation:

σₓ = σ / √n

where σ = 4.0 and n = 225

So

μₓ = μ = 77.4

σₓ = σ / √n = 4.0 / √225 = 4.0 / 15 = 0.3

Hope this helps!

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.

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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.