Suppose a population is known to be normally distributed with a mean, μ, equal to 144 and a standard deviation, σ, equal to 27. Approximately what percent of the population would be between 144 and 171?

34.13% of the population would be between 144 and 171

The z score is used to determine by how many standard deviations the raw score is above or below the mean. The z score is given by:

Given that μ = 144, σ = 27:

For x = 144:

Therefore, From the normal distribution table: P(144 < x < 171) = P(0 < z < 1) = P(z < 1) - P(z < 0) = 0.8413 - 0.5 = 34.13%

Hence 34.13% of the population would be between 144 and 171

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Approximately 34%.

Step-by-step explanation:

Calculate the z-scores:

For 144;  z  = 144 - 144 / 27 = 0

For 172: z = 171-144/27 = 1.

From the Normal distribution table = 0 gives  0.5000 and 1 gives 0.8413.

So required percentage = 0.8413 - 0.5000 = 0.3413.

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