Question

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.

Answer 1

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.

Answer 2

`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`.