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Determine the 90 th percentile of the distribution, which is \(N(65,25)\).

Short Answer

Expert verified
The 90th percentile of the distribution \(N(65,25)\) is approximately 97.

Step by step solution

01

Identify the relevant parameters of the normal distribution

The mean \(\mu\) is 65 and the standard deviation \(\sigma\) is 25 for the normal distribution \(N(65,25)\).
02

Use the standard normal distribution table

Refer to the standard normal distribution (also known as the Z-table). The value closest to 0.9000 is 1.28. This is the z-score that represents the 90th percentile.
03

Use the z-score formula to find the raw score (x)

Use the formula \(x = \mu + z*\sigma\), where \(\mu\) is the mean, \(\sigma\) is the standard deviation and z is the z-score. Plug in the known values to get \(x = 65 + 1.28*25\).

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