ONKAR SINGH

2022-10-14

Q3. Let X be a normal distribution random variable with mean 9 and variance 16. Find the 25th percentile of X.

Eliza Beth13

To find the 25th percentile of a normal distribution random variable, we can use the properties of the standard normal distribution.
Given that X is a normal distribution random variable with a mean ($\mu$) of 9 and a variance (${\sigma }^{2}$) of 16, we need to standardize X to the standard normal distribution with a mean of 0 and a standard deviation of 1.
The standardization formula is:
$Z=\frac{X-\mu }{\sigma }$
where Z represents the standardized random variable.
In this case, we have:
$Z=\frac{X-9}{4}$
Now, we need to find the value of Z that corresponds to the 25th percentile, which can be denoted as ${Z}_{0.25}$.
Using a standard normal distribution table or a calculator, we can find that ${Z}_{0.25}$ is approximately -0.6745.
To find the corresponding value of X, we can rearrange the standardization formula:
$X=Z·\sigma +\mu$
Substituting the values:
$X=\left(-0.6745\right)·4+9$
Simplifying:
$X=-2.698+9$
$X=6.302$
Therefore, the 25th percentile of X is approximately 6.302.

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