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Suppose X is normally distributed with mean 15 and standard deviation 5.

Find the 90th percentile for X.
19.5
19.1
21.4
20

User Malou
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1 Answer

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Final answer:

The 90th percentile of a normally distributed variable X with a mean of 15 and a standard deviation of 5 is found by converting to a standard normal distribution and using the Z-value that corresponds to a cumulative probability of 0.90. The percentile value is calculated to be approximately 21.4.

Step-by-step explanation:

To find the 90th percentile of a normally distributed random variable X with a mean (μ) of 15 and a standard deviation (σ) of 5, you would typically use a standard normal distribution table or a calculator. The 90th percentile is the value which 90 percent of the distribution falls below.

We will first convert our normal variable to the standard normal variable Z, which has a mean of 0 and a standard deviation of 1, using the formula Z = (X - μ) / σ. Then we will find the Z value that corresponds to 0.90 in the cumulative distribution function of the standard normal distribution.

After looking up 0.90 in a Z-table or using a calculator, we find that the Z value is approximately 1.28. Then, we can find the specific value for X using the formula X = μ + Zσ = 15 + 1.28(5).

By calculating this, we will get:
X = 15 + 6.4 = 21.4.

Therefore, the 90th percentile for X is approximately 21.4.

User Tore Rudberg
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