Suppose a population is known to be normally distributed with a mean, μ, equal to 116 and a standard deviation, σ, equal to 14. Approximately what percent of the population would be between 88 and 130?

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altavistard altavistard · Answer 1 · 2021-04-27T01:53:22+02:00

Answer:

81.9% of the population would be between 88 and 130.

Step-by-step explanation:

Many of today's calculators have built-in probability distribution functions.

Here, on a TI -83 calculator, we need the function normalcdf(.

Type in the following: normalcdf(88,130,116, 14)

The result is the desired probabiity: 0.819, which is equivalent to 81.9%.

81.9% of the population would be between 88 and 130.

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raphealnwobi raphealnwobi · Answer 2 · 2021-11-23T10:40:24+01:00

81.85% of the population would be between 88 and 130

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 = 116, σ = 14:

For x = 88:

[tex]z=\frac{88-116}{14} =-2\\\\\\For\ x=130:\\\\z=\frac{130-116}{14} =1[/tex]

Therefore, From the normal distribution table: P(88 < x < 130) = P(-2 < z < 1) = P(z < 1) - P(z < -2) = 0.8413 - 0.0228 = 81.85%

Hence 81.85% of the population would be between 88 and 130

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