Question

Suppose a normal distribution has a mean of 62 and a standard deviation of

Answer 1

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:

$z = \frac{x-\mu}{\sigma} \\\\where\ x=raw\ score,\mu=mean,\sigma=standard\ deviation$

Given that μ = 62, σ = 4

For x = 54:

$z=\frac{54-62}{4}=-2$

For x = 66:

$z=\frac{66-62}{4}=1$

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