Answer:
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.
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:
[tex]z=\frac{x-\mu}{\sigma} \\\\where\ \mu=mean,x=raw\ score,\sigma=standard\ deviation[/tex]
Given that μ = 144, σ = 27:
For x = 144:
[tex]z=\frac{144-144}{27} =0\\\\\\For\ x=171:\\\\z=\frac{171-144}{27} =1[/tex]
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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