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Suppose a normal distribution has a mean of 62 and a standard deviation of
4. What is the probability that a data value is between 54 and 66?
A.79.9%
B.77.9%
C.81.9%
D.78.9%

Sagot :

raphealnwobi

The probability that a data value is between 54 and 66 is 81.9%

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\ x=raw\ score,\mu=mean,\sigma=standard\ deviation[/tex]

Given that μ = 62, σ = 4

For x = 54:

[tex]z=\frac{54-62}{4}=-2[/tex]

For x = 66:

[tex]z=\frac{66-62}{4}=1[/tex]

P(54 < x < 66) = P(-2 < z < 1) = P(z < 1) - P(z < -2) = 0.8413 - 0.0228 = 81.9%

The probability that a data value is between 54 and 66 is 81.9%

Find out more on z score at: brainly.com/question/25638875

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