A personality test is administered to a large group of subjects. Five scores are shown below, in original units and in standard units. Fill in the bla. [79645272−1.80.8−−−1.4]

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A personality test is administered to a large group of subjects. Five scores are shown below, in original units and in standard units. Fill in the bla.  [79645272−1.80.8−−−1.4]

Nadine Salcido · Accepted Answer

The normal standard variable is z.z=x−μσWhen&amp;nbsp;x=79,z=1.8i.e.1.8=79−μσ⇒μ+1.8σ=78When&amp;nbsp;x=64,z=0.8i.e.0.8=64−μσ⇒μ+0.8σ=64Solving the two equations above results inσ=14Now,&amp;nbsp;μ+1.8(14)=78⇒μ=52.8When&amp;nbsp;x=52z=52−52.814=−0.06When&amp;nbsp;z=−1.4−1.4=x−52.814⇒=33.2Answer:&amp;nbsp;μ=52.8&amp;nbsp;&amp;nbsp;and&amp;nbsp;&amp;nbsp;σ=14

usaho4w · Accepted Answer

Use z-score formula to find the missing values.   z=x−meansd   sd= standard deviation   First you have to figure out the mean and standard deviation.   Lets

A personality test is administered to a large group of

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