A random sample of size 16 is to be taken from a normal population having mean of 100 andvariance 4. What is the 90th percentile of the distribution of x?

The 90th percentile of the distribution of x is 100.64.

Given that,

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]

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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".

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