Respuesta :
The 90th percentile of the distribution of x is 100.64.
Given that,
- A random sample of size 16 is to be taken from a normal population having mean of 100 and variance 4
calculation:
Here we have
[tex]n=16,\mu=100,\sigma^{2}=4[/tex]
Now we requried z-score that has 0.90 area to its left. z-score 1.28 has 0.90 area to its left.
So required 90th percentile is
[tex]z=\frac{\bar{x}-\mu}{\sigma/\sqrt{n}}\\\\1.28=\frac{\bar{x}-100}{2/\sqrt{16}}\\\\\bar{x}=100.64[/tex]
Learn more about the population here: https://brainly.com/question/905400?referrer=searchResults
The distribution value is "100.64".
Given:
[tex]\to n=16\\\\\to \mu=100\\\\ \to \sigma^{2}=4[/tex]
To find:
distribution [tex]\bar{x}=?[/tex]
Solution:
We require a z-score with such a 0.90 area to its left in this case. The area to the left of the z-score 1.28 is 0.90. As a consequence, the required 90th percentile is
Using formula:
[tex]\to z=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}}\\\\[/tex]
[tex]\to 1.28=\frac{\bar{x}-100}{\frac{2}{\sqrt{16}}}\\\\\to 1.28=\frac{\bar{x}-100}{\frac{2}{\sqrt{4^2}}}\\\\\to 1.28=\frac{\bar{x}-100}{\frac{2}{4}}\\\\\to 1.28=\frac{\bar{x}-100}{\frac{1}{2}}\\\\\to 1.28=\frac{\bar{x}-100}{0.5}\\\\\to 0.64 =\bar{x}-100\\\\\to \bar{x}=100+0.64\\\\\to \bar{x}=100.64\\\\[/tex]
Therefore, the answer is "100.64".
Learn more about the distribution here:
brainly.com/question/10137107
