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On a standardized exam, the scores are normally distributed with a mean of 200 and a standard deviation of 10. Find the z-score of a person who scored 181 on the exam.

User Felixqk
by
7.7k points

2 Answers

2 votes

Answer:2.3

Explanation:

On a standardized exam, the scores are normally distributed with a mean of 200 and a standard deviation of 40. Find the z-score of a person who scored 292 on the exam.

\text{Draw a picture:}

Draw a picture:

μ−3σ

μ−2σ

μ−σ

μ

μ+σ

μ+2σ

μ+3σ

80

120

160

200

240

280

320

292

z=\frac{x-\mu}{\sigma}

z=

σ

x−μ

z-score formula

z=\frac{292-200}{40}

z=

40

292−200

Plug in values

z=\frac{92}{40}

z=

40

92

Subtract

z=2.3

z=2.3

Divide

\text{They scored 2.3 standard deviations above the mean.}

They scored 2.3 standard deviations above the mean.

User Some
by
8.5k points
3 votes

Answer:


z=-1.9

Explanation:

We have been that on a standardized exam, the scores are normally distributed with a mean of 200 and a standard deviation of 10. We are asked to find the z-score of a person who scored 181 on the exam.

We will use z-score formula to solve our given problem.


z=(x-\mu)/(\sigma), where,

z = z-score,

x = Random sample score,


\mu = Mean,


\sigma = Standard deviation.

Upon substituting our given values in above formula, we will get:


z=(181-200)/(10)


z=(-19)/(10)


z=-1.9

Therefore, the z-score of the person would be
-1.9.

User Jil Jung Juk
by
7.6k points

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