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

Question

Eliza Beth13 · Accepted Answer

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 (

μ

) of 9 and a variance (

σ^{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 - μ}{σ}

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 \cdot σ + μ

Substituting the values:

X = (- 0.6745) \cdot 4 + 9

Simplifying:

X = - 2.698 + 9

X = 6.302

Therefore, the 25th percentile of X is approximately 6.302.

Q3. Let X be a normal distribution random

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