Question

a variable x has a normal distribution with mean 54 and standard deviation 5. approximately what percentage of the observations will fall between 44 and 64?

Answer 1

For a normal distribution with mean 54 and standard deviation 5 95.4% of them fall between 44 and 64

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:

$z=\frac{x-\mu}{\sigma} \\\\where\ \mu=mean.\sigma=standard\ deviation, x=raw\ score\\\\Given\ that\ mean(\mu)=54,standard \ deviation (\sigma)=5\\\\For\ x=44:\\\\z=\frac{44-54}{5}=-2\\\\For\ x=64:\\\\z=\frac{64-54}{5}=2$

Therefore., from the normal distribution table:

P(44<x<64) = P(-2< z < 2) = P(z < 2) - P(z < -2) = 0.9772 - 0.0228 = 0.9544

Hence, About 95.4% of them fall between 44 and 64.

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