Answer:
Step-by-step explanation:
Given that:
population mean = 10
variance [tex]\sigma^2[/tex] = 25 ; [tex]\sigma[/tex] =[tex]\sqrt{ 25}[/tex] = 5
sample size n = 28
The standard deviation of the sample mean:
= [tex]{\dfrac{5}{\sqrt{28}}[/tex]
= 0.95
To halve the standard deviation of the sample mean, the size of how large the sample will be can be computed as follows:
[tex]\dfrac{sd_1}{sd_2} = \dfrac{\dfrac{\sigma}{\sqrt{n}} }{\dfrac{\sigma}{\sqrt{n}} }[/tex]
[tex]\implies \dfrac{\dfrac{5}{\sqrt{28}} }{\dfrac{5}{\sqrt{28}} }[/tex]
[tex]= \dfrac{5}{\sqrt{28}} \times \dfrac{\sqrt{28}}{5}[/tex]
= 2
n = 4 × 28
n = 112
The standard deviation is 0.95 and if the standard deviation is halved then the sample mean is 112.
It is the measure of the dispersion of statistical data. Dispersion is the extent to which the value is in a variation.
A normal population has a mean of 10 and a variance of 25.
A random sample of sizes n-28 is selected.
A. The standard deviation will be
[tex]\sigma = \dfrac{\sqrt{Var(x)}}{n} \\\\\sigma = \dfrac{\sqrt{25}}{\sqrt{28}} \\\\\sigma = 0.95[/tex]
B. The sample be if you want to halve the standard deviation of the sample mean
[tex]\rm \dfrac{SD_1}{SD_2} = 2 = \dfrac{\sqrt{n}}{\sqrt{28}}\\\\n = 4*28 \\\\n= 112[/tex]
More about the standard deviation link is given below.
https://brainly.com/question/12402189
